Class 10 Maths - Pair Of Linear Equations In Two Variables

Solve the following equations graphically: $3x-2y = 4$ , $5x-2y = 0$

$3x-2y = 4$ ..(i)

$2y = 3x-4$

$y = (3x-4)/2$
When

$x = 0,$

$y = (3×0-4)/2 = (0-4)/2 = -4/2 = -2$
when

$x = 2,$

$y = (3×2-4)/2 = (6-4)/2 = 2/2 = 1$
when

$x = 4,$

$y = (3×4-4)/2 = (12-4)/2 = 8/2 = 4$

$x=$	$0$	$,2$	$,4$
$y=$	$-2$	$,1$	$,4$

Plot the above points on graph. Join them.

$5x-2y = 0$ …(ii)

$2y = 5x$

$y = 5x/2$
When

$x = 0,$

$y = 0$
When

$x = 2,$

$y = 5×2/2 = 5$
When

$x = -2,$

$y = 5×-2/2 = -5$

$x=$	$0$	$,2$	$,-2$
$y=$	$0$	$,5$	$,-5$

Plot the above points on graph. Join them.
It is clear from the graph that the two lines intersect at

$(-2,-5).$
So the solution of the given equations are

$x = -2$ and

$y = -5.$

1725052_1812125_ans_546a3816d5aa48dfa3b89085381dfbc5.PNG

Solve the following pairs of linear equations by substition method;
$3x - y = 3$
$9x - 3y = 9$

Given equations are

$3x - y = 3$ ...(i)

$9x - 3y = 9$ ....(ii)
from

$eq^n$ (i),y = 3x - 3 ....(iii)
On substituting y = 3x - 3 in

$eq^n$ (ii) we get

$\Rightarrow 9x - 3(3x - 3) = 9$

$\Rightarrow 9x - 9x + 9 = 9$

$\Rightarrow 9 = 9$
This equality is true for all values of x, therefore given pair of equation have infinity many solutions..

Solve the following pair of linear equations by substitution method:
$x + y = 2m$
$mx - ny = m^2+n^2$

Given equations are

$x + y = 2m$ .....(i)

$mx - ny = m^2 n^2$ ....(ii)

from

$eq^n$ (i), x = 2m-y ....(iii)

On substituting x = 2m - y in

$eq^n$ (ii) we get

$\Rightarrow m(2m - y) - ny = m^2 + n^2$

$\Rightarrow 2m^2 - my - ny = m^2 + n^2$

$\Rightarrow -y (m+n) - ny = m^2 - 2m + n^2$

$\Rightarrow -y (m+n) = -m^2 - 2m^2 + n^2$

$\Rightarrow y = \dfrac{n^2 - m^2}{-(a + b)}(\therefore .(a^2 - b^2) = (a -b)(a - b))$

$\Rightarrow y = \dfrac{(n - m(n + m))}{-(n + m)}$

$y = - (n-m) = m - n$

Now on putting

$y = m - n$ in

$eq^n(iii)$ , we get

$\Rightarrow x = 2m - (m - n)$

$\Rightarrow = m + n$

Thus,

$x = m$ and

$y =m - n$ is the required solution.

Solve the following pair of linear equations by substitution method:
$\dfrac{x}{2} + y = 0.8$
$x + \dfrac{y}{2} = \dfrac{7}{10}$

Given equations are

$\dfrac{x}{2} + y = 0.8$ .....(i)

$x + \dfrac{y}{2} = \dfrac{7}{10}$ ....(ii)

from

$eq^n$ (i)can be written as,

$\Rightarrow \dfrac{x}{2} + y =0.8$ ,

$\Rightarrow \left(\dfrac{x+2y}{2}\right) \Rightarrow x + 2y = 0.8 \times 2$

$\Rightarrow x + 2y = 1.6$ ....(iii)

On substituting

$x = 1.6 - 2y$ in

$eq^n$ (ii) we get

$\Rightarrow 1.6 - 2y + \dfrac{y}{2} = \dfrac{7}{10}$

$\Rightarrow \dfrac{3.2 - 4y + y}{2} =\dfrac{7}{10}$

$\Rightarrow 3.2 - 3y = \dfrac{7}{10} \times 2$

$\Rightarrow 3.2-3y = \dfrac{7}{10} \times 2$

$\Rightarrow -3y = \dfrac{7}{10}\times 2 - 3.2$

$\Rightarrow -3y = \dfrac{14}{10} - \dfrac{32}{10}$

$\Rightarrow -3y = -\dfrac{18}{10}$

$\Rightarrow y=\dfrac{18}{10}\times \dfrac{1}{3}$

$\Rightarrow y = \dfrac{6}{10} = 0.6$

Now on putting

$y = 0.6$ in

$eq^n(iii)$ , we get

$\Rightarrow x + 2y = 16)$

$\Rightarrow x + 2(0.6) = 1.6$

$\Rightarrow x + 1.2 = 1.6$

$\Rightarrow x = 0.4$

Thus,

$x = 0.4$ and

$y =0.6$ is the required solution.

Solve the following pair of linear equations by substitution method:
$\dfrac{x}{a} +\dfrac{y}{b} = 2 , a\neq 0,b \neq 0$
$ax - by = a^2 - b^2$

Given equations are

$\dfrac{x}{a} +\dfrac{y}{b} = 2 , a\neq 0,b \neq 0$ .....(i)

$ax - by = a^2 - b^2$ .....(ii)

from

$eq^n$ (i)can be written as,

$\Rightarrow \dfrac{x}{a}+\dfrac{t}{b} = 2$ ,

$\Rightarrow \left(\dfrac{bx + ay}{ab}\right)=2$

$\Rightarrow bx + ay = ab \times 2$

$\Rightarrow bx + ay = 2ab$

$\Rightarrow x = \dfrac{2ab - ay}{b}$ ...(iii)

On substituting x = \dfrac{2ab - ay}{b} in

$eq^n$ (ii) we get

$\Rightarrow a\left(\dfrac{2ab - ay}{b}\right) - by = a^2 - b^2$

$\Rightarrow \dfrac{2a^2b - a^2y}{b} -by = a^2 - b^2$

$\Rightarrow \dfrac{2a^2b - a^2y - b^2y}{b} = a^2 - b^2$

$\Rightarrow 2a^2b - a^2y - b^2y = b(a^2 - b^2)$

$\Rightarrow 2a^2b - y(a^2+b^2) = a^2b - b^3$

$\Rightarrow -y(a^2 + b^2) = a^2 b - b_3 - 2a^2b$

$\Rightarrow -y(a^2 + b^2 = - b(a^2 + b^2)))$

$\Rightarrow y=\dfrac{-b(a^2 + b^2)}{-(a^2+b^2)}= b$

Now on putting

$y = b$ in

$eq^n(iii)$ , we get

$\Rightarrow x = \dfrac{2ab - ab}{b}$

$\Rightarrow x = a$

Thus,

$x = a$ and

$y =b$ is the required solution.

Solve graphically the following system of linear equations if it has unique solution :
$3x + y = 2$
$6x + 2y = 1$

The given pair of linear equations is

$3x + y = 2$ or

$3x + y - 2 = 0$

$6x + 2y = 1$ or

$6x + 2y - 1 = 0$
On comparing the given equations with standard form of pair of linear equations i.e

$a_1 x + b_1 y + c_1 = 0$ and

$a_2 x + b_2 y + c_2 = 0$ , we get

$a_1 = 3, b_1 = 1$ and

$c_1 = -2$
and

$a_2 = 6, b_2 = 2$ and

$c_2 = -1$

$\dfrac{a_1} {a_2} = \dfrac{3} {6} = \dfrac{1} {2}, \dfrac{b_1} {b_2} = \dfrac{1} {2} = 1$ and

$\dfrac{c_1} {c_2} = \dfrac{-2} {-1} = 2$

$\therefore \dfrac{a_1} {a_2} = \dfrac{b_1} {b_2} \neq \dfrac{c_1} {c_2}$
The lines representing the given pair of linear equations are parallel.

Find graphically the co-ordinates of the vertices of the triangle formed by the lines $y = 0, y = x$ and $2x + 3y= 10$ . Hence find the area of the triangle formed by these lines.

$y = 0 \quad ........(i)$
The graph of

$y = 0$ is the

$X$ axis.

$y = x \quad ........(ii)$
When

$x = 1, y = 1.$
When

$x = 2, y = 2.$
When

$x = 3, y = 3.$

$x=$	$1$	$,2$	$,3$
$y=$	$1$	$,2$	$,3$

Mark the above points on graph. Join them.

$2x+3y = 10 \quad ....(iii)$

$3y = 10-2x$

$y = \dfrac{10-2x}{3}$
When

$x = 0.5, y = \dfrac{(10-2×0.5)}{3} = \dfrac{10-1}{3} = \dfrac{9}{3} = 3$

When

$x = 2, y = \dfrac{(10-2×2)}{3} = \dfrac{10-4}{3} = \dfrac{6}{3} = 2$

When

$x = 5, y = \dfrac{10-2×5}{3} = \dfrac{10-10}{3} = 0$

$x=$	$0.5$	$,2$	$,5$
$y=$	$3$	$,2$	$,0$

Mark the above points on graph. Join them.
From the graph, it is clear that the vertices of the triangle formed by the lines are

$A(2,2), B(5,0)$ and

$C(0,0)$ .
Area of triangle formed by these lines

$=\dfrac{1}{2}\times base\times height$

$\Rightarrow \dfrac{1}{2}\times 5\times 2$

$\Rightarrow 5$ square units
Hence area of the triangle is

$5$ square units.

1722693_1812184_ans_0ed5f3096a8742afabfd060271fc100d.PNG

Solve the following pair of linear equations by substitution method:
0.5x - 0.8y = 3.4
0.6x + 0.3y = 0.3

Given equations are

$0.5 + 0.8y = 3.4$ ...(i)

$0.6x - 0.3 = 0.3$ ....(ii)
from

$eq^n..... (ii), 2x - y = 1$

$y = 2x - 1 ....(iii)$
On substituting

$y = 2x -1$ in

$eq^n$ (i) we get

$\Rightarrow 0.5x + 0.8(2x - 1) = 3.4$

$\Rightarrow0.5x + 1.6 - 0.8 = 3.4$

$\Rightarrow 2.1x = 3.4 + 0.8$

$\Rightarrow 21x = 4.2$

$x = \dfrac {4.2}{2.1} = 2$
Now on putting

$x = 2$ in

$eq^n(iii)$ , we get

$\Rightarrow y = 2(2) - 1$

$y = 4 - 1$

$\Rightarrow y = 3$
Thus,

$x = 2$ and

$y = 3$ is the required solution.

Solve the following pair of linear equations by substitution method:
$S - t = 3$
$\dfrac{s}{3} + \dfrac{t}{2} = 6$

Given equations are

$s - t = 3$ ...(i)

$\dfrac{s}{3} + \dfrac{t}{2} = 6$ ....(ii)
from

$eq^n (i), s = 3 + t ....(iii)$
On substituting

$s = 3 +t$ in

$eq^n$ (ii) we get

$\Rightarrow \dfrac{3 + t}{3}+ \dfrac{t}{2}= 6$

$\Rightarrow \dfrac{2(3+t) + 3t}{6}= 6$

$\Rightarrow 6 + 2t +3t = 6 \times 6$

$\Rightarrow 5t = 36 -6$

$\Rightarrow t = \dfrac{30}{5} = 6$
Now on putting

$t = 6$ in

$eq^n(iii)$ , we get

$\Rightarrow s = 3 + 6$

$\Rightarrow s = 9$
Thus,

$s = 9$ and

$t = 6$ is the required solution.

Using the same axes of co-ordinates and the same unit, solve graphically.
$x+y = 0, 3x 2y = 10$

$x+y = 0$ ..(i)

$y = -x$
When

$x = -3,$

$y = 3$
When

$x = -2,$

$y = 2$
When

$x = -1,$

$y = 1$

$x=$	$-3$	$,-2$	$,-1$
$y=$	$3$	$,2$	$,1$

Mark the above points on graph. Join them.

$3x-2y = 10$ ..(ii)

$3x= 2y+10$

$x = (2y+10)/3$
When

$y = 1$

$x = (2×1+10)/3 = 12/3 = 4$
When

$y = -2$

$x = (2×-2+10)/3 = 6/3 = 2$
When

$y = 4$

$x = (2×4+10)/3 = 18/3 = 6$

$x=$	$4$	$,2$	$,6$
$y=$	$1$	$,-2$	$,4$

Mark the above points on graph. Join them.
It is clear from the graph that the two lines intersect at

$(2,-2).$
So the solution of the given equations are

$x = 2$ and

$y = -2$ .
1725106_1812162_ans_706d38baac994a41ba8aaef5fd9ffc14.PNG

Solve the following pair of linear equations by substitution method:
$7x - 15y = 2$
$x + 2y = 3$

Given equations are

$7x - 15y = 2 ...(i)$

$x + 2y = 3 ....(ii)$
from

$eq^n ....(ii),$

$x = 3 - 2y ....(iii)$
On substituting x= 3 -2y in

$eq^n$ (i) we get

$\Rightarrow 7(3-2y) - 15y = 2$

$\Rightarrow 21 - 14y - 15y = 2$

$\Rightarrow 21 - 29y = 2$

$\Rightarrow -29y = -19$

$y = \dfrac{19}{29}$
Now on putting

$y = \dfrac{19}{29}$ in

$eq^n(iii)$ , we get

$\Rightarrow x = 3- 2 \left( \dfrac{19}{29}\right)$

$\Rightarrow x = 3 - \dfrac{38}{29}$

$\Rightarrow x = \dfrac{87-38}{29}$

$\Rightarrow x = \dfrac{49}{29}$
Thus,

$x = \dfrac{49}{29}$ and

$y = \dfrac{19}{29}$ is the required solution.

Solve the following systems equations by eliminations method;
$3x - 5y - 4 = 0...(i)$
$9x = 2y + 7....(ii)$

Given pair of linear equations is

$3x - 5y - 4 = 0...(i)$
And

$9x = 2y + 7....(ii)$
On multiplying eq. (i) by 3 to make the coefficients of x equal, we get the equations as

$9x - 15y - 12 = 0....(iii)$
On subtracting Eq.(ii) from Eq.(iii) we get

$9x - 15y - 12 - 9x = 0 - 2y - 7$

$\Rightarrow - 15y + 2y = -7 + 12$

$\Rightarrow -13y = 5$

$\Rightarrow y = -\dfrac{5}{13}$
On putting

$y = - \dfrac{5}{13}$ in Eq. (ii) we get

$9x = 2\left(-\dfrac{5}{13}\right) + 7$

$\Rightarrow 9x = -\dfrac{10}{13} + 7$

$\Rightarrow 9x = \dfrac{-10 + 91}{13}$

$\Rightarrow 9x = \dfrac{81}{13}$

$\Rightarrow x \dfrac{81}{13 \times 9}$

$\Rightarrow x = \dfrac{9}{13}$
Hence

$x = \dfrac{9}{13}$ and

$-\dfrac{5}{13}$ which is the required solutions

Solve the following systems equations by eliminations method;
$8x + 5y = 9$
$3x + 2y = 4$

Given pair of linear equations is

$8x + 5y = 9...(i)$
And

$3x + 2y = 4....(ii)$
On multiplying eq. (i) by 2 and Eq. (ii) by 5 to make the coefficients of x equal, we get the equations as

$16x + 10y = 18....(iii)$

$15x + 10y = 20.....(iv)$
On subtracting Eq.(iii) from Eq.(iv) we get

$15x + 10y - 16x - 10y = 20 - 18$

$\Rightarrow - x = 2$

$\Rightarrow x = -2$
On putting

$x = -2$ in Eq. (ii) we get

$3x + 2y = 4$

$\Rightarrow 3(-2) + 2y = 4$

$\Rightarrow -6 + 2y = 4$

$\Rightarrow 2y = 4 +6$

$\Rightarrow 2y = 10$

$\Rightarrow y = \dfrac{10}{2} = 5$
Hence

$x = -2$ and

$y = 5$ which is the required solutions

Solve the following systems equations by eliminations method;
$0.4x + 1.5y = 6.5$
$0.3x - 0.2y = 0.9$

Given pair of linear equations is

$0.4x + 1.5y = 6.5...(i)$

And

$0.3x - 0.2y = 0.9....(ii)$

On multiplying eq. (i) by 3 and Eq. (ii) by 4 to make the coefficients of x equal, we get the equations as

$1.2x + 4.5y = 19.5....(iii)$

$1.2x - 0.8y = 3.6.....(iv)$

On subtracting Eq.(iii) from Eq.(iv) we get

$1.2x - 0.8y - 1.2x - 4.5y = 3.6 - 19.5$

$\Rightarrow -5.3y =-15.9$

$\Rightarrow y= -\dfrac{15.9}{5.3}$

$\Rightarrow y = 3$

On putting

$y = 3$ in Eq. (ii) we get

$0.3x + 0.2y = 0.9$

$\Rightarrow 0.3x - 0.2(3) = 0.9$

$\Rightarrow 0.3x - 0.6 = 0.9$

$\Rightarrow 0.3x = 1.5$

$\Rightarrow x = 5$

Hence

$x = 5$ and

$y = 3$ which is the required solutions

Solve the following systems equations by eliminations method;
$\sqrt{2}x - \sqrt{3}y = 0$
$\sqrt{5}x + \sqrt{.2}y = 0$

Given pair of linear equations is

$\sqrt{2}x + \sqrt{3}y = 0...(i)$
And

$\sqrt{5}x + \sqrt{2}y = 0....(ii)$
On multiplying eq. (i) by

$\sqrt{2}$ and Eq. (ii) by

$\sqrt{3}$ to make the coefficients of x equal, we get the equations as

$2x - \sqrt{6}y = 0....(iii)$

$\sqrt{15}x + \sqrt{6}y = 0.....(iv)$
On adding Eq.(iii) from Eq.(iv) we get

$2x - \sqrt{6}y - \sqrt{15}x + \sqrt{6}y =0$

$\Rightarrow 2x + \sqrt{15}x = 0$

$\Rightarrow x(2 + \sqrt{15}) = 0$

$\Rightarrow x = 0$
On putting

$x = 0$ in Eq. (ii) we get

$\sqrt{2}x + \sqrt{3}y = 0$

$\Rightarrow \sqrt{2}(0) + \sqrt{3}y = 0$

$\Rightarrow \sqrt{3}y = 0$

$\Rightarrow y =0$
Hence

$x = 0$ and

$y = 0$ which is the required solutions

Solve the following systems equations by eliminations method;
$2x + 3y = 8$
$4x + 6y =7$

Given pair of linear equations is

$2x + 3y = 8...(i)$
And

$4x + 6y = 7....(ii)$
On multiplying eq. (i) by 2 to make the coefficients of x equal, we get the equations as

$4x + 6y = 16....(iii)$
On subtracting Eq.(ii) from Eq.(iii) we get

$4x + 6y - 4x - 6y = 16 - 7$

$\Rightarrow 0 = 9$
Which is a false equation involving no variable.
so, the given equation has no solution i.e this pair of linear equation is inconsistent

Solve the following pair of linear equations by substitution method:
$\dfrac{bx}{a} + \dfrac{ay}{b} = a^2 + b^2$
$x + y = 2ab$

Given equation are

$\dfrac{bx}{a} + \dfrac{ay}{b} = a^2 + b^2$ ....(i)

$x + y = 2ab$ ....(ii)

$eq^n(i)$ can be written as,

$\Rightarrow \dfrac{bx}{a}+ \dfrac{ay}{b} = a^2 +b^2$

$\Rightarrow b^2x + a^2 y = ab \times (a^2 +b^2)$

$\Rightarrow x = \dfrac{ab\times(a^2 + b^2) - a^2 y}{b^2}$ ....(iii)

Now from

$eq^n$ (ii), y = 2ab - x...(iv)

On substituting y = 2ab - x in

$eq^n$ (iii), we get

$\Rightarrow x = \dfrac{a^3b + b^3a - 2a^3b + a^2x}{b^2}$

$\Rightarrow b^2x = b^3a - a^3 b + a^2x$

$\Rightarrow b62 x - a^2x = b^3a - a^3b$

$\Rightarrow (b^2-a^2)x = ab(b^2 - a^2)$

$\Rightarrow x = ab$

Now putting x = ab in

$eq^n$ (iv) we get

$\Rightarrow y = 2ab - ab$

$\Rightarrow y = ab$

Thus,

$x = ab$ and

$y = ab$ is the required solution

Solve graphically the following system of linear equations if is has unique solution:
$2x - 6y + 10 = 0$
$3x - 9y + 15 =0$

The given pair of linear equations is

$2x - 6y + 10 = 0$

$3x - 9y + 15 =0$

On comparing the given equations with standard form of pair of linear

equations i.e

$a_1 x + b_1 y + c_1 = 0$ and

$a_2 x + b_2 y + c_2 = 0$ , we get

$a_1 = 2, b_1 = -6$ and

$c_1 = 10$

and

$a_2 = 3, b_2 = -9$ and

$c_2 = 15$

$$\frac{a_1} {a_2} = \frac{2} {3} , \frac{b_1} {b_2} =

\frac{-6} {-9} = \frac{2} {3}

$and$ \frac{c_1} {c_2} = \frac{10} {15} = \frac{2} {3}$$

$$\therefore \frac{a_1} {a_2} = \frac{b_1} {b_2} = \frac{c_1}

{c_2}$$

$\therefore$ The lines representing the given pair of linear equations will coincide.

Solve the following systems equations by eliminations method;
$3x + 4y - 10...(i)$
$2x - 2y = 2....(ii)$

Given pair of linear equations is

$3x + 4y = 10 ...(i)$
And

$2x - 2y = 7....(ii)$
On multiplying eq. (i) by 2 and Eq.(ii) by 3 to make the coefficients of x equal, we get the equations as

$6x + 8y = 20....(iii)$

$6x - 6y = 6 ....(iv)$
On subtracting Eq.(iii) from Eq.(iv) we get

$6x - 6y - 6x - 8y = 6 - 20$

$\Rightarrow - 14y = -14$

$\Rightarrow y =1$
On putting

$y = 1$ in Eq. (ii) we get

$2x -2(1) = 2$

$\Rightarrow 2x - 2 = 2$

$\Rightarrow x = 2$
Hence

$x = 2$ and

$y = 1$ which is the required

Solve the following systems equations by eliminations method;
$2x + 3y = 46$
$3x + 5y = 74$

Given pair of linear equations is

$2x + 3y = 46...(i)$
And

$3x + 5y = 74....(ii)$
On multiplying eq. (i) by 3 and Eq. (ii) by 2 to make the coefficients of x equal, we get the equations as

$6x + 9y = 138....(iii)$

$6x + 10y = 148.....(iv)$
On subtracting Eq.(iii) from Eq.(iv) we get

$6x + 10y - 6x - 9y = 148 - 138$

$\Rightarrow y = 10$
On putting

$y = 10$ in Eq. (ii) we get

$3x + 5y = 74$

$\Rightarrow 3x + 5(10) = 74$

$\Rightarrow 3x + 50 = 74$

$\Rightarrow 3x = 74 - 50$

$\Rightarrow 3x = 24$

$\Rightarrow x = 8$
Hence

$x = 8$ and

$y = 10$ which is the required solutions

Solve the following systems equations by eliminations method;
$x + y = 5$
$2x - 3y = 4$

Given pair of linear equations is

$x + y = 5...(i)$
And

$2x = 3y = 4....(ii)$
On multiplying eq. (i) by 2 to make the coefficients of x equal, we get the equations as

$2x + 2y = 1 0....(iii)$
On subtracting Eq.(ii) from Eq.(iii) we get

$2x + 2y - 2x + 3y = 10 - 4$

$\Rightarrow 5y = 6$

$\Rightarrow y= \dfrac{6}{5}$
On putting

$y = \dfrac{6}{5}$ in Eq. (i) we get

$x + y = 5$

$\Rightarrow x + \dfrac{6}{5} = 5$

$\Rightarrow x = \dfrac{25 - 6}{5}$

$\Rightarrow x = \dfrac{19}{5}$
Hence

$x = \dfrac{19}{5}$ and

$\dfrac{6}{5}$ which is the required solutions

The sum of two numbers is $16$ and the sum of their reciprocals is $1 / 3 .$ Find the numbers.

Let the two numbers be x and y.
According to the question,

$x+y=16 \ldots(i)$

$\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{3} \ldots(\text { ii })$
Eq. (ii) can be re - written as

$\dfrac{y+x}{x y}=\dfrac{1}{3} \ldots(\text { iii })$
On putting the value of

$x+y=16$ in Eq. (iii), we get

$\dfrac{16}{x y}=\dfrac{1}{3}$

$\Rightarrow x y=48$

$\Rightarrow x=\dfrac{48}{y}$
On putting the value of

${\mathrm{x}}=\dfrac{48}{\mathrm{y}}$ in Eq. (i), we get

$\dfrac{48}{y}+y=16$

$\Rightarrow 48+y^{2}=16 y$

$\Rightarrow y^{2}-16 y+48=0$

$\Rightarrow \mathrm{y}^{2}-12 \mathrm{y}-4 \mathrm{y}+48=0$

$\Rightarrow y(y-12)-4(y-12)=0$

$\Rightarrow(y-4)(y-12)=0$

$\Rightarrow \mathrm{y}=4$ and 12
If

$y=4,$ then

$x=\dfrac{48}{4}=12$
If

$y=12,$ then

$x=\dfrac{48}{12}=4$
Hence, the two numbers are

$4$ and

$12 .$

Solve the following equations by elimination method
$\dfrac{1}{x} - \dfrac{1}{y} = 1$
$\dfrac{1}{x} + \dfrac{1}{y} = 7.x \neq 0.y \neq 0$

Given pair of linear equations is

$\dfrac{1}{x} - \dfrac{1}{y} = 1..(i)$

$\dfrac{1}{x} + \dfrac{1}{y} = 7...(ii)$
Adding Eq.(i) and Eq.(ii), we get

$\dfrac{1}{x} - \dfrac{1}{y} + \dfrac{1}{x} + \dfrac{1}{y} = 1 + 7$

$\Rightarrow \dfrac{1}{x} + \dfrac{1}{x} = 8$

$\Rightarrow \dfrac{2}{8} = x$

$\Rightarrow x= \dfrac{1}{4}$
On putting

$x = \dfrac{1}{4}$ in Eq. (ii) we get

$\dfrac{1}{x} + \dfrac{1}{y} = 7$

$\Rightarrow \dfrac{1}{\dfrac{1}{4}} + \dfrac{1}{y} = 7$

$4 + \dfrac{1}{y} = 7 \Rightarrow \dfrac{1}{y} = 7 - 4 = 3$

$y = \dfrac{1}{3}$
Hence,

$x = \dfrac{1}{4}$ and

$\dfrac{1}{3}$ , which is required solutions

Two positive numbers differ by 3 and their product isFind the numbers.

Let the two numbers be

$x$ and

$y$

Given

$:\ x-y=3$

$...(i)$

And,

$x \times y=54$

$\Rightarrow {x}=\dfrac{54}{y}$

$...(ii)$

On putting the value of

$x=\dfrac{54}{y}$ in equation

$(i)$

$\Rightarrow \dfrac{54}{y}-y=3$

$\Rightarrow 54-y^{2}=3 y$

$\Rightarrow y^{2}+3 y-54=0$

$\Rightarrow y^{2}+9 y-6 y-54=0$

$\Rightarrow y(y+9)-6(y+9)=0$

$\Rightarrow(y-6)(y+9)=0$

$\Rightarrow y=-9$ and

$6$

Given that the numbers are positive.

$\therefore y=6$

and

$x=\dfrac{54}{6}=9$

Hence the two numbers are

$9$ and

$6$

Solve the following equations bu elimination method:
$99x + 101y = 499$
$101x + 99y = 501$

Given pairs of linear equations is

$99x + 101y = 499...(i)$
And

$101x + 99y = 501....(ii)$
Given pair of linear equations is
On multipying Eq. (i) by

$101$ and Eq (ii) by

$99$ to make the coefficients of x equal, we get the equation as

$9999x + 10201y = 50399...(iii)$

$9999x + 9801y = 49599$
On subtracting Eq.(iii) and Eq.(iv), we get

$\Rightarrow 9999x + 9801y - 9999x - 10201y = 49599 - 50399$

$\Rightarrow 9801y - 10201y = 49599 - 50399$

$-400y = -800$

$\Rightarrow y = \dfrac{800}{400}$

$\Rightarrow y = 2$
On putting

$y=2$ in Eq.(ii) we get

$99x + 101(2) = 499 \Rightarrow 99x + 202 = 499$

$\Rightarrow 99x = 297$

$\Rightarrow x = 297/99$

$\Rightarrow x = 3$
Hence,

$x = 3$ and

$y = 2$ , which is the required solutions.

Solve the following equations bu elimination method:
$29x - 23y = 110$
$23x - 29y = 98$

Given pairs of linear equations is

$29x - 23y = 110...(i)$
And

$23x - 29y = 98....(ii)$
Given pair of linear equations is
On multipying Eq. (i) by 23 and Eq (ii) by 29 to make the coefficients of x equal, we get the equation as

$667x - 529y = 2530...(iii)$

$667x - 841y = 2842....(iv)$
On subtracting Eq.(iii) and Eq.(iv), we get

$\Rightarrow 667x - 841y - 667x + 529y = 2842 - 2530$

$\Rightarrow -312y = 312$

$\Rightarrow y = -1$
On putting

$y=-1$ in Eq.(ii) we get

$29x - 23(-1) = 110 \Rightarrow 29x + 23 = 110$

$\Rightarrow 29x = 110 - 23$

$\Rightarrow 29x = 87$

$\Rightarrow x = 3$
Hence,

$x = 3$ and

$y = -1$ , which is the required solutions.

Solve the following systems of equations by elimination method:
$\dfrac{x}{6} + \dfrac{y}{15}= 4 , \dfrac{x}{3} - \dfrac{y}{12} =\dfrac{19}{4}$

Given pair of linear equations is

$\dfrac{x}{6} + \dfrac{y}{15}= 4$

And

$\dfrac{x}{3} - \dfrac{y}{12} =\dfrac{19}{4}$

On multipying Eq. (ii) by

$\dfrac{1}{2}$ to make the coefficients of x equal, we get the equation as

$\dfrac{x}{6} - \dfrac{y}{24} = \dfrac{19}{8}...(iii)$

On Subtracting Eq.(ii) and Eq.(iii), we get

$\dfrac{x}{6}-\dfrac{y}{24}- \dfrac{x}{6} - \dfrac{y}{15}-=\dfrac{19}{8} - 4$

$\Rightarrow - \dfrac{y}{24} - \dfrac{y}{15} = \dfrac{19-32}{8}$

$\Rightarrow \dfrac{5y - 8y}{120} = \dfrac{19 - 32}{8}$

$\Rightarrow \dfrac{-13y}{120} = \dfrac{-13}{8}$

$y = \dfrac{13}{8} \times \dfrac{120}{13}$

$\Rightarrow y = 15$

On putting

$y=15$ in Eq.(ii) we get

$\dfrac{x}{6} + \dfrac{y}{15} = 4$

$\dfrac{x}{6} + \dfrac{15}{15} = 4$

$\dfrac{x}{6} = 4 - 1$

$\dfrac{x}{6} = 3$

$x = 18$

Hence,

$x = 18$ and

$y = 15$ , which is the required solutions.

Solve the following system of equations by elimination method:
$x + \dfrac{6}{y} = 6 ,3x - \dfrac{8}{y} = 5$

Given pairs of linear equations is

$x + \dfrac{6}{y} = 6....(i)$

$3x - \dfrac{8}{y} = 5...(ii)$
Given pair of linear equations is
On multipying Eq. (i) by 3 to make the coefficients of x equal, we get the equation as

$3x - \dfrac{18}{y} = 18...(iii)$
On subtracting Eq.(ii) and Eq.(Iii), we get

$3x + \dfrac{18}{y} - 3x +\dfrac{8}{y} = 18 - 5$

$\Rightarrow \dfrac{26}{y} = 13$

$\Rightarrow y = 2$
On putting

$y=2$ in Eq.(ii) we get

$x + \dfrac{6}{y} = 6$

$\Rightarrow x + \dfrac{6}{2} = 6$

$\Rightarrow x = 3$
Hence,

$x = 3$ and

$y = 2$ , which is the required solutions.

Solve the following system of equations by elimination method:
$2x + 5y = 1$
$2x + 3y = 3$

Given pair of linear equations is

$2x + 5y = 1...(i)$
And,

$2x + 3y = 3....(ii)$

On subtracting Eq.(i) from Eq.(ii) we get

$(2x + 3y) - (2x + 5y) = 3 - 1$

$\Rightarrow 2x + 3y - 2x - 5y = 3 - 1$

$\Rightarrow -2y = 2$

$\Rightarrow y = -1$

On putting

$y = -1$ in Eq. (ii) we get

$2x + 3(-1) = 3$

$\Rightarrow 2x - 3 = 3$

$\Rightarrow 2x = 6$

$\Rightarrow x = \dfrac{6}{2}$

$\Rightarrow x = 3$

Hence, the required solution is

$x = 3$ and

$y = -1$ .

Solve the following equations bu elimination method:
$217x + 131y = 913$
And $131x + 217y = 827$

Given pairs of linear equations is

$217x + 131y = 913...(i)$

And

$131x + 217y = 827....(ii)$

Given pair of linear equations is

On multipying Eq. (i) by 131 and Eq (ii) by 217 to make the coefficients of x equal, we get the equation as

$28427x + 17161y = 119603...(iii)$

$28427x + 47089y = 179459$

On subtracting Eq.(iii) and Eq.(iv), we get

$\Rightarrow 28427x + 47089y - 28427x - 17161y = 179459 - 119603$

$\Rightarrow 47089y -17161y = 179459 - 119603$

$29928y = 59856$

$\Rightarrow y = \dfrac{59856}{29928}$

$\Rightarrow y = 2$

On putting

$y=2$ in Eq.(ii) we get

$131x + 217(2) = 117 \Rightarrow 131x + 434 = 827$

$\Rightarrow 131x = 393$

$\Rightarrow = \dfrac{393}{131}$

$\Rightarrow x = 3$

Hence,

$x = 3$ and

$y = 2$ , which is the required solutions.

Solve the following systems of equations by elimination method:
$3x - \dfrac{8}{y} = 5$
$x - \dfrac{y}{3} = 3$

Given pair of linear equations is

$\dfrac{x}{2}+\dfrac{2y}{3} = -1...(i)$

And

$x - \dfrac{y}{3} = 3...(ii)$

On multipying Eq. (ii) by 2 to make the coefficients of y equal, we get the equation as

$2x - \dfrac{2y}{3} = 6...(iii)$

On adding Eq.(i) and Eq.(ii), we get

$\dfrac{x}{2}+\dfrac{2y}{3}+ x - \dfrac{2y}{3} = -1 + 6$

$\Rightarrow \dfrac{x}{2} + 2x = 5$

$\Rightarrow \dfrac{x + 4x}{2} = 5$

$\Rightarrow \dfrac{5x}{2} = 5$

$\Rightarrow x = 2$

On putting

$x=2$ in Eq.(ii) we get

$x - \dfrac{y}{3} = 3$

$2 - \dfrac{y}{3} = 3$

$-\dfrac{y}{3} = 3 - 2$

$-\dfrac{y}{3} = 1$

$y = -3$

Hence,

$x = 2$ and

$y = -3$ , which is the required solutions.

Solve the following equations by elimination method
$\dfrac{4}{x} + \dfrac{7}{y} = 11, \dfrac{3}{x} - \dfrac{5}{y} = -2\ \ (x = \neq 0.y \neq 0)$

Given pair of linear equations is

$\dfrac{4}{x} + \dfrac{7}{y} = 11..(i)$

$\dfrac{3}{x} - \dfrac{5}{y} = -2...(ii)$

On multiplying Eq.(i) by 3 and Eq. (ii) by 4 to make the coefficients of

$^x equal$ , we get the equations

$\dfrac{12}{x} + \dfrac{21}{y} = 33...(iii)$

$\dfrac{12}{x} - \dfrac{20}{y} = -8...(iv)$

On subtracting Eq.(i) and Eq.(ii), we get

$\dfrac{12}{x} + \dfrac{20}{y} - \dfrac{12}{x} - \dfrac{21}{y} = -8 - 33$

$\Rightarrow \dfrac{20}{y} - \dfrac{21}{x} = -41$

$\Rightarrow -\dfrac{41}{y} = -41$

$\Rightarrow \dfrac{41}{41} = y$

$\Rightarrow y = 1$

On putting

$y = 1$ in Eq. (ii) we get

$\dfrac{3}{x} - \dfrac{5}{y} = -2$

$\Rightarrow \dfrac{3}{x} - 5 = -2$

$4 + \dfrac{3}{x} = -2 + 5 \Rightarrow \dfrac{3}{x} = 3$

$x = 1$

Hence,

$x = \dfrac{1}{3}$ and

$a\dfrac{1}{2}$ , which is required

Two numbers differ by 2 and their product is $360 .$ Find the numbers.

Let the two numbers be

$x$ and

$y$

Given

$:\ x-y=2$

$...(i)$

And

$x \times y=360$

$\Rightarrow x=\dfrac{360}{y}$

$\ldots({ii})$

On putting the value of

${x}=\dfrac{360}{y}$ in equation

$(i),$ we get

$\Rightarrow \dfrac{360}{y}-y=2$

$\Rightarrow 360-y^{2}=2 y$

$\Rightarrow y^{2}+2 y-360=0$

$\Rightarrow y^2+20y-18y-360=0$

$\Rightarrow y(y+20)-18(y+20)=0$

$\Rightarrow(y-18)(y+20)=0$

$\Rightarrow y=-20$ and

$18$

Given that the numbers are positive

$\Rightarrow y=18,$ then

$x=\dfrac{360}{18}=20$

Hence, the two numbers are

$20$ and

$18$ .

Two numbers differ by 4 and their product is 192 . Find the numbers.

Let the two numbers be

$\mathrm{x}$ and

$\mathrm{y}$ .
According to the question,

$x-y=4 \ldots(\text { i })$
Also,

$x \times y=192$

$\Rightarrow x=\dfrac{192}{y} \ldots(\text { ii })$
On putting the value of

${\mathrm{x}}=\dfrac{192}{\mathrm{y}}$ in Eq. (i), we get

$\dfrac{192}{y}-y=4$

$\Rightarrow 192-y^{2}=4 y$

$\Rightarrow y^{2}+4 y-192=0$

$\Rightarrow \mathrm{y}^{2}+16 \mathrm{y}-12 \mathrm{y}-192=0$

$\Rightarrow \mathrm{y}(\mathrm{y}+16)-12(\mathrm{y}+16)=0$

$\Rightarrow(y-12)(y+16)=0$

$\Rightarrow \mathrm{y}=-16$ and

$12$
But

$\mathrm{y}=-16$ can't be the one number as it is given that the numbers are positive.

$\Rightarrow \mathrm{y}=12,$ then

$\mathrm{x}=\dfrac{192}{12}=16$
Hence, the two numbers are

$16$ and

$12$ .

Solve the following equations by elimination method
$x = y = 2xy$ ,
$x - y = 6xy$

Given pair of linear equations is

$x + y = 2xy....(i)$ ,

$x - y = 6xy...(ii)$
On adding Eq. (i) and Eq(ii), we get

$x + y + x - y = 2xy + 6xy$

$\Rightarrow 2x = 8xy$

$\Rightarrow \dfrac{2x}{8x} = y$

$\Rightarrow y = \dfrac{1}{4}$ and

$x = 0$
On putting

$y = \dfrac{1}{4}$ in Eq. (ii), we get

$x - \dfrac{1}{4} = 6xy$

$\Rightarrow \dfrac{24x-1}{4} = 6x\left(\dfrac{1}{4}\right)$

$\Rightarrow 4x - 1 = 6x \Rightarrow-1 = 6x - 4x$

$\Rightarrow -1x = 2x$

$\Rightarrow x = -\dfrac{1}{2}$
On putting

$x = 0$ we get

$y = 0$
Hence

$x = -\dfrac{1}{2},0$ and

$y = \dfrac{1}{4},0$ which is the required

Solve the following equations by elimination method
$5x + y = 19xy , 7x - 2y = 8xy$

Given pair of linear equations is

$5x + 3y = 19xy....(i)$ ,

$7x - 2y = 8xy...(ii)$
On multiplying Eq.(i) by 2 and Eq.(ii) by 3 to make the coefficients of y equal, we get the equation as

$10x + 6y = 38xy....(iii)$

$21x - 6y = 24xy...(iv)$
On adding Eq. (i) and Eq(ii), we get

$10x + 6y + 21x - 6y = 38xy + 24xy$

$\Rightarrow 31x = 62xy$

$\Rightarrow \dfrac{31x}{62x} = y$

$\Rightarrow y = \dfrac{1}{2}$ and

$x = 0$
On putting

$y = \dfrac{1}{2}$ in Eq. (ii), we get

$7x - 2y = 8xy$

$\Rightarrow 7x - 2\left(\dfrac{1}{2}\right) = 8x\left(\dfrac{1}{2}\right)$

$\Rightarrow 7x - 1 = 4x \Rightarrow-1 = 4x - 7x$

$\Rightarrow -1 = -3x$

$\Rightarrow x = \dfrac{1}{3}$
On putting

$x = 0$ we get

$y = 0$
Hence

$x = \dfrac{1}{3},0$ and

$y = \dfrac{1}{2},0$ which is the required solution

Solve the following equations by elimination method
$\dfrac{3a}{x} - \dfrac{2b}{y} + 5 = 0 , \dfrac{a}{x} - \dfrac{3b}{y} - 2 = 0(x = \neq 0.y \neq 0)$

Given pair of linear equations is

$\dfrac{3a}{x} + \dfrac{2b}{y} = -5..(i)$

$\dfrac{a}{x} - \dfrac{3b}{y} = 2...(ii)$

On multiplying Eq.(ii) by 3 to make the coefficients of

$^x equal$ , we get the equations

$\dfrac{3a}{x} - \dfrac{9b}{y} = 6...(iii)$

On subtracting Eq.(iii) from Eq.(i), we get

$\dfrac{3a}{x} - \dfrac{9b}{y} - \dfrac{3a}{x} - \dfrac{2b}{y} = 6-( - 5)$

$\Rightarrow - \dfrac{9b}{y} - \dfrac{2b}{y} = 11$

$\Rightarrow -\dfrac{11b}{y} = 11$

$\Rightarrow y = -b$

On putting

$y = b$ in Eq. (ii) we get

$\dfrac{a}{x} - \dfrac{3b}{y} = 2$

$\Rightarrow \dfrac{a}{x} + \dfrac{3b}{y} = 2$

$\dfrac{a}{x} = -2 +3 \Rightarrow \dfrac{a}{x} = 1$

$x = a$

Hence,

$x = a$ and y = -b$$, which is required

Solve the following equations by elimination method
$x + y = 7xy$ ,
$2x - 3y = -xy$

Given pair of linear equations is

$x + y = 7xy....(i)$ ,

$2x - 3y = -xy...(ii)$

On multiplying Eq.(i) by 2 to make the coefficients of y equal, we get the equation as

$2x + 2y = 14xy....(iii)$

On subtracting Eq. (ii) from Eq(iii), we get

$2x + 2y - 2x + 3y = 14xy + xy$

$\Rightarrow 2y + 3y = 15xy$

$\Rightarrow 5y = 15xy$

$\Rightarrow \dfrac{5y}{15y} = x$

$\Rightarrow x = \dfrac{1}{3} and = 0$

On putting

$x = \dfrac{1}{3}$ in Eq. (ii), we get

$2x - 3y = xy$

$\Rightarrow 2\left(\dfrac{1}{3}\right) - 3y = 6\left(\dfrac{1}{2}\right)y$

$\Rightarrow \left(\dfrac{2}{3}\right) - 3y = 2y \Rightarrow\left( \dfrac{2}{3}\right)-= 5y$

$\Rightarrow y = \dfrac{2}{15}$

On putting

$x = 0$ we get

$y = 0$

Hence

$x = -\dfrac{1}{3},0$ and

$y = \dfrac{2}{15},0$ which is the required

A number consisting of two digits is seven times the sum of its digits. When 27 is subtracted from the number, the digits are reversed. Find the number.

${\textbf{Step -1: Let us suppose a two digit number }}$ $\mathbf{10x + y.}$

${\text{where unit place digit is }}$ $y$ ${\text{and tens place digit is}}$ $x.$

${\text{It is given that the two digit number is seven times of sum of its digits}}{\text{. }}$

${\text{i}}{\text{.e}}{\text{.}}$ $10x + y = 7\left( {x + y} \right)$

$\Rightarrow 10x + y = 7x + 7y$

$\Rightarrow 10x - 7x = 7y - y = 0$

$\Rightarrow 3x = 6y$

$\Rightarrow$ $x = 2y \ldots \left( 1 \right)$

${\textbf{Step -2: Finding the Required Number}}$

${\text{It is given that, if 27 is subtracted from the number }}$

${\text{then the resultant number will be reversed of the original 2 digit number.}}$

${\text{i}}{\text{.e}}{\text{.}}$ $10x + y - 27 = 10y + x$

$\Rightarrow 10x - x = 10y - y + 27$

$\Rightarrow 9x - 9y = 27$

$\Rightarrow x - y = 3 \ldots \left( 2 \right)$

${\text{Substitute the value of}}$ $x$ ${\text{from equation }}\left( 1 \right)$ ${\text{into equation }}\left( 2 \right)$

$\Rightarrow 2y - y = 3$

$\Rightarrow y = 3$

${\text{Now, substitute the value of }}$ $y$ ${\text{in equation }}\left( 1 \right)$

$\Rightarrow x = 2 \times 3$

$\Rightarrow x = 6$

${\text{So, the required number is}}$ $10\left( 6 \right) + 3 = 63.$

${\textbf{Hence, the two digit number is 63.}}$

Solve the following equations by elimination method
$\dfrac{2x + 5y}{xy} = 6 , \dfrac{4x - 5y}{xy} = -3(where x = \neq 0 and y \neq 0)$

Given pair of linear equations is

$\dfrac{2x + 5y}{xy} = 6....(i)$ ,

$\dfrac{4x - 5y}{xy} = -3$
Or

$4x - 5y = -3xy....(ii)$
On adding Eq. (i) and Eq(ii), we get

$2x + 5y + 4x - 5y = 6xy - 3xy$

$\Rightarrow 6x = 3xy$

$\Rightarrow \dfrac{6x}{3x} = y$

$\Rightarrow y = 2 and x = 0$
On putting

$y = 2$ in Eq. (ii), we get

$2x + 5(2) = 6xy$

$\Rightarrow 2x + 10 = 6x(2)$

$\Rightarrow 2x + 10 = 12x \Rightarrow 2x - 12x = -10$

$\Rightarrow -10x = -10$

$\Rightarrow x = 1$
On putting

$x = 0$ we get

$y = 0$
Hence

$x = 0,1$ and

$y = 0,2$ which is the required solution

Solve the following equations by elimination method
$\dfrac{2}{x} + \dfrac{3}{y} = 13$
$\dfrac{5}{x} - \dfrac{4}{y} = 2.x \neq 0.y \neq 0$

Given pair of linear equations is

$\dfrac{2}{x} + \dfrac{3}{y} = 13..(i)$

$\dfrac{5}{x} - \dfrac{4}{y} = -2...(ii)$
On multiplying Eq.(i) by 5 and Eq. (ii) by 2 to make the coefficients of

$x$ equal, we get the equations

$\dfrac{10}{x} + \dfrac{15}{y} = 65...(iii)$

$\dfrac{10}{x} - \dfrac{9}{y} = -4...(iv)$
On subtracting Eq.(i) and Eq.(ii), we get

$\dfrac{10}{x} - \dfrac{8}{y} - \dfrac{10}{x} - \dfrac{15}{y} = -4 - 65$

$\Rightarrow -\dfrac{8}{y} - \dfrac{15}{y} = -69$

$\Rightarrow \dfrac{-23}{-69} = y$

$\Rightarrow \dfrac{23}{69} = y$

$\Rightarrow y = \dfrac{1}{2}$
On putting

$y = \dfrac{1}{2}$ in Eq. (ii) we get

$\dfrac{5}{x} - \dfrac{4}{y} = -2$

$\Rightarrow \dfrac{5}{x} - \dfrac{4}{\dfrac{1}{2}} = 7$

$\dfrac{5}{x} - 8 = 7 \Rightarrow \dfrac{5}{x} = 15$

$x = \dfrac{1}{3}$
Hence,

$x = \dfrac{1}{3}$ and

$y=\dfrac{1}{2}$ , which is required

The age of the father is 3 years more than 3 times the son's age. 3 years here, the age of the father will be 10 years more than twice the age of the son. Find their present ages.

Let the age of father

$=x$ years
And the age of his son

$=y$ years
According to the question,

$x=3+3 y \ldots(i)$
Three year here, Father's age

$=(\mathrm{x}+3)$ years
Son's age

$=(\mathrm{y}+3)$ years
According to the question,

$(x+3)=10+2(y+3)$

$\Rightarrow x+3=10+2 y+6$

$\Rightarrow x=2 y+13 \ldots(\text { ii })$
From Eq. (i) and (ii), we get

$3+3 y=13+2 y$

$\Rightarrow 3 y-2 y=13-3$

$\Rightarrow \mathrm{y}=1 \mathrm{0}$
On putting the value of

$y=7$ in Eq. (i),
we get

$x=3+3(10)$

$\Rightarrow x=3+30$

$\Rightarrow x=33$
Hence, the age of father is

$33$ years and the age of his son is

$10$ yrs.

For which value of a, the following system of linear equations has no solutions :
$ax + 3y = a - 2, 12x + ay = a$

Given, pair of equations

$ax + 3y = a - 2$ and

$12x + ay = a$
On comparing the given equation with standard form
i.e.

$a_1x + b_1y + c_1 = 0$ and

$a_2x + b_2y + c_2 = 0$ ,
we get

$a_1 = a, b_1 = 3$ and

$c_1 = - \left(a - 2\right)$
and

$a_2 = 12, b_2 = a$ and

$c_2 = - a$
For infinitely many solutions,

$\dfrac{a_1} {a_2} = \dfrac{b_1} {b_2} \neq \dfrac{c_1} {c_2}$

$\dfrac{a} {12} = \dfrac{3} {a} \neq \dfrac{- \left(a - 2 \right)} {-a}$

$\therefore$ I II III
On taking I and II terms, we get

$\dfrac{a} {12} = \dfrac{3} {a}$

$\Rightarrow a^{2} = 36$

$\Rightarrow a = \sqrt{36}$

$\Rightarrow a = \pm 6$

Ten years hence, a man's age will be twice the age of his son. Ten years ago, the man was four times as old as his son. Find their present ages

Let the age of a man = x years And the age of his son = y years
Ten years hence,
Man's age

$=(\mathrm{x}+10)$ years
Son's age

$=(\mathrm{y}+10)$ years
According to the question,

$(x+10)=2(y+10)$

$\Rightarrow x+10=2 y+20$

$\Rightarrow x=2 y+20-10$

$\Rightarrow x=2 y+10 \ldots(i)$
Ten years ago,
Father's age

$=(x-10)$ years
Son's age

$=(\mathrm{y}-10)$ years
According to the question,

$(x-10)=4(y-10)$

$\Rightarrow x-10=4 y-40$

$\Rightarrow \mathrm{x}=4 \mathrm{y}-30 \ldots(\mathrm{ii})$
From Eq. (i) and (ii), we get

$2 y+10=4 y-30$

$\Rightarrow 2 y-4 y=-30-10$

$\Rightarrow-2 y=-40$

$\Rightarrow y=20$
On putting the value of

$y=20$ in Eq. (i), we get

$x=2 y+10$

$\Rightarrow x=2(20)+10$

$\Rightarrow x=50$
Hence the age of the man is

$50$ yrs and the age of his son is

$20$ yrs.

Form the pairs of linear equations for the following problems and find their solutions by elimination method:
Aftab tells his daughter,"seven years ago, I was seven times as old as you were then, Also, three years from now I shall be three times as old as you will be.'Find their present ages.

Let the presents age of father i.e Aftab

$= x$ yr
And the present age of his daughter

$= y$ yr
Seven years ago,
Aftab age

$= (x - 7)$ yr
Daughter age

$= (y - 7)$ yr
According to the question

$(x - 7) = 7(y - 7)$

$\Rightarrow x - 7 = 7y - 49$

$\Rightarrow x - 7y = -42...(i)$
After three years
Aftab age

$= (x + 3)$ yr
Daughter age

$= (y + 3)$ yr
According to the question

$(x + 3) = 3(y + 3)$

$\Rightarrow x + 3 = 3y +9$

$x - 3y = 6...(ii)$
Now we can solve this by an elimination method
On subtracting Eq. (ii) from (i) we get

$x - 3y - x + 7y = 6 - (-42)$

$\Rightarrow-3y + 7y = 6 + 42$

$\Rightarrow 4y = 48$

$\Rightarrow y = 12$
On putting

$y = 12$ in Eq.(ii) we get

$x - 3(12) = 6$

$\Rightarrow x - 36 = 6$

$\Rightarrow x =42$
Hence, the age of aftab is

$42$ years and age of his daughter is

$12$ years

The sum of the digits of a two-digit number is $15 .$ The number obtained by interchanging the digits exceeds the given number by $9 .$ Find the number.

Let unit's digit

$=y$ and the ten's digit

$=x$
So, the original number

$=10 x+y$
The sum of the number

$=10 x+y$
The sum of the digit

$=x+y$
According to the question,

$x+y=15 \ldots(i)$
After interchanging the digits, the number

$=x+10 y$ and

$10 x+y+9=x+10 y$

$\Rightarrow 10 x+y+9=x+10 y$

$\Rightarrow 10 x-x+y-10 y=-9$

$\Rightarrow 9 x-9 y=-9$

$\Rightarrow x-y=-1 \ldots(\text { ii })$
On adding Eq. (i) and (ii), we get

$x+y+x-y=15-1$

$\Rightarrow 2 x=14$

$\Rightarrow \mathrm{x}=7$
On substituting the value of

$x=5$ in Eq. (i), we get

$x+y=15$

$\Rightarrow 7+y=15$

$\Rightarrow \mathrm{y}=8$
So, the Original number

$=10 x+y$

$=10 \times 7+8$

$=70+8$

$=78$
Hence, the two digit number is

$78$ .

Check whether the following equations are consistent or inconsistent, solve them graphically.
$5x - 3y = 11$
$-10x + 6 y = -22$

Equations : -

$5x - 3y = 11 ..... (i)$

$-10x + 6y = -22 ..... (ii)$

From (i) and (ii)

$\dfrac {a_{1}}{a_{2}} = \dfrac {5}{-10} = \dfrac {-1}{2}$

$\dfrac {b_{1}}{b_{2}} = \dfrac {-3}{6} = \dfrac{-1}{2}$

$\dfrac {c_{1}}{c_{2}} = \dfrac {11}{-22} = \dfrac {-1}{2}$ ,

$\therefore \dfrac {a_{1}}{a_{2}} = \dfrac {b_{1}}{b_{2}} = \dfrac {c_{1}}{c_{2}}$

$\Rightarrow$ The linear equation are consistent and the lines are coincident

Hence they have infinitely many solutions .

Since, both lines (i) and (ii) are coincident, Hence represent the same line.

$5x - 3y = 11 \Rightarrow y = \dfrac {5x - 11}{3} .... (\star)$

(1) put

$x = 0$ in

$(\star) \Rightarrow y = \dfrac {-11}{3}$

(2) Put

$y = 0$ in

$(\star) \Rightarrow 5x - 11 = 0\Rightarrow x = \dfrac {11}{5}$

$x$	$0$	$\dfrac {11}{5}$
$y$	$\dfrac {-11}{3}$	$0$

2109412_1836160_ans_87e3626bd58b4bb6bbc2fa0a8f205f76.jpg

The sum of two digit number and the number obtained by reversing the digit is 66 . If the digit of the number differ by 2 . Find the number . How many such numbers are there?

Let the unit place digit be x, the tens place digit be y and value of the number be 10y + x from the problem(10y+x) + ( 10x +y) = 66 and x - y =2
11x + 11 y = 66 and x - y = 2

$\Rightarrow x+y = 6 and x - y = 2$
Solve the equation, we get

$2x = 8 \Rightarrow x =\dfrac{8}{2}\Rightarrow 4$
Substitute x value in the equation x + y = 6 , we get

$4+y = 6\Rightarrow y = 6-4=2$

Substitute x , y values in the equation(10y+x) and (10x+y)10(2) +4 and 10 (4) +2

$\Rightarrow 20+4=24 and 40+2=42$

$\therefore$ The number is 24 or 12 Thus we have two such numbers

Form the pairs of linear equations for the following problems and find their solutions by elimination method:
Five years ago, Nuri was thrice as old as sonu. Ten years later, Nuri will be twice as old as sonu. How old are Nuri and Sonu?

Let the presents age of father i.e Nuri = x yr

And the present age of Sonu = y yr

Five years ago, age of Nauri age = (x - 5)yr

Sonu age = (y - 5)yr

According to the question

$(x + 5) = 3(y + 5)$

$\Rightarrow x - 5 = 3y +15$

$x - 3y = 10...(i)$

After ten years

Nuri age =(y + 10)yr

sonu age =(y + 10)yr

According the question

$(x + 10) = 2(y + 10)$

$\Rightarrow x + 10 = 2y +20$

$\Rightarrow x - 2y = 10...(ii)$

Now we can solve this by an elimination method

On subtracting Eq. (ii) from (i) we get

$x - 2y - x + 3y = 10 - (-10)$

$\Rightarrow-2y + 3y = 10 + 10$

$\Rightarrow y = 20$

On putting y = 20 in Eq.(i) we get

$x - 3(20) = -10$

$\Rightarrow x - 60 = -10$

$\Rightarrow x =50$

Hence, the age of Nuri is 50years and age of his daughter is 20 years

Form the pairs of linear equations for the following problems and find their solutions by elimination method:
The difference between two numbers is 26 and one number is three times the other find them

Let the one number = x

And the other number = y

According to the question

$x - y = 26$

and

$x = 3y$

$x - 3y = 0...(ii)$

Now we can solve this by an elimination method

On subtracting Eq. (ii) from (i) we get

$x - 3y - x + y = 0 - 26$

$\Rightarrow-3y + y = -26$

$\Rightarrow -2y = -26$

$\Rightarrow y = 13$

On putting y = 13 in Eq.(ii) we get

$x - 3(13) = 0$

$\Rightarrow x - 39 = 0$

$\Rightarrow x =39$

Hence, the other number are 39 and 13.

Five years ago, A was thrice as old as B and ten years later, A shall be twice as old as B. What are the present ages of A and B?

Let the age of

$A=x$ years And the age of

$B=y$ years
Five years ago,
A's age

$=(\mathrm{x}-5)$ years

$\mathrm{B}^{\prime} \mathrm{s}$ age

$=(\mathrm{y}-5)$ years
According to the question,

$(x-5)=3(y-5)$

$\Rightarrow x-5=3 y-15$

$\Rightarrow \mathrm{x}=3 \mathrm{y}-10 \ldots(\mathrm{i})$
Ten years later,
A's age

$=(x+10)$ B's age

$=(y+10)$
According to the question,

$(x+10)=2(y+10)$

$\Rightarrow x+10=2 y+20$

$\Rightarrow x=2 y+10 \ldots(\text { ii })$
From Eq. (i) and (ii), we get

$3 y-10=2 y+10$

$\Rightarrow 3 y-2 y=10+10$

$\Rightarrow y=20$
On putting the value of

$y=20$ in Eq. (i), we get

$x=3(20)-10$

$\Rightarrow x=50$
Hence, the age of person A is

$50$ years and Age of B is

$20$ years.

Solve each of the following pairs of equation by reducing then to a pair of linear equations.
$\dfrac{x+y}{xy}=2,\dfrac{x-y}{xy}=6$

$\dfrac{x+y}{xy}=2,\dfrac{x-y}{xy}=6$

$\dfrac{x}{xy}+\frac{y}{xy}=2,\dfrac{x}{xy}-\dfrac{y}{xy}=6$

$\dfrac{1}{y}+\dfrac{1}{x}=2,\dfrac{1}{y}-\dfrac{1}{x}=6$

$Let \dfrac{1}{x}=aand \dfrac{1}{y}=b, then$

$a+b = 2 ...(1),-a+b=6...(2)$

$Equations (1)+(2), gives$

$2b = 8\Rightarrow b =\dfrac{8}{2}=4$ Substitute

$b = 4$ in equation (1), gives

$a+4=2\Rightarrow a=2-4=-2$ But

$a = \dfrac{1}{x}\Rightarrow -2=\dfrac{1}{x}\Rightarrow x=-\dfrac{1}{2}$

$b = \dfrac{1}{y}\Rightarrow 4\dfrac{1}{y}\Rightarrow y=\dfrac{1}{4}$

$Solutions (x,y)=\left ( -\dfrac{1}{2},\dfrac{1}{4} \right )$

Check whether the following equations are consistent or inconsistent , solve them graphically.
2x -2y -2 = 0 and 4x - 4y -5 =0

2x -2y -2 = 0 and 4x - 4y -5 =0

$\dfrac{a_{1}}{a_{2}} = \dfrac{2}{4}=\dfrac{1}{2} ,\dfrac{b_{1}}{b_{2}}=\dfrac{-2}{-4}=\dfrac{1}{2} =\dfrac{c_{1}}{c_{2}}=\dfrac{-2}{-5}=\dfrac{2}{5}$

$\therefore \dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} \neq \dfrac{c_{1}}{c_{2}}$

$\therefore$ the linear equation are consistent

$\therefore$ They have infinity solutions.

2x - 2y-2 = 0 4x - 4y -5 =0

x y (x ,y) x y (x,y)

0 -1 (0,-1) 2

$\frac{3}{4}$

1796645_1836203_ans_6386cf4460ca42b2a369a19e535e9281.png

Solve each of the following pairs of equation by reducing then to a pair of linear equations.
$\dfrac{1}{3x+y}+\dfrac{1}{3x-y}=\dfrac{3}{4},\dfrac{1}{2(3x+y)}-\dfrac{1}{2(3x-y)}=\dfrac{-1}{8}$

$\dfrac{1}{3x+y}+\dfrac{1}{3x-y}=\dfrac{3}{4},\dfrac{1}{2(3x+y)}-\dfrac{1}{2(3x-y)}=\dfrac{-1}{8}$

$Let\dfrac{1}{3x+y}=a$

$and \dfrac{1}{3x-y}=b, then$

$a+b = \dfrac{3}{4}\Rightarrow 4a+4b=3........(1)\dfrac{a}{2}=\dfrac{-1}{8}\Rightarrow 4a=-1......(2)$

$Equation (3)+(4),Gives$

$6x=6\Rightarrow x=\dfrac{6}{6}=1$

$Substitute x = 1 in equation (3), given$

$3(1)+y=4\Rightarrow y=4-3=1$

$Solution (x,y)=(1,1)$

Check whether the following equations are consistent or inconsistent , solve them graphically.
2x + y - 6 = 0 and 4x - 2y -4 = 0

2x + y - 6 = 0 and 4x - 2y -4 = 0

$\dfrac{a_{1}}{a_{2}} = \dfrac{2}{4}=\dfrac{1}{2} ,\dfrac{b_{1}}{b_{2}}=\dfrac{1}{-2} \dfrac{c_{1}}{c_{2}}=\dfrac{-6}{-4}=\dfrac{3}{2}$

$\therefore \dfrac{a_{1}}{a_{2}} \neq \dfrac{b_{1}}{b_{2}}$

$\therefore$ the linear equation are consistent

$\therefore$ They have infinity solutions.

2x + y = 6 4x - 2y = 4

x y (x ,y) x y (x,y)

3 0 (0,6) 0 2 (0,-2)

1796693_1836191_ans_3bf8316d9b594adcbda19f990c544636.png

For solving each pair of equation, in this exercise, use the method of elimination by equating coefficients:

$\frac{x-y}{6}=2(4-x)$
$2x+y=3(x-4)$

The given pair of linear equations are

$\frac { x-y}{6 } = 2(4-x)$

$\Rightarrow 13x -y = 48 ...............(i)$ [ on simplifying ]

$2x +y = 3(x -4)$

$\Rightarrow x -y = 12 ...............(ii)$ [ on simplifying]
Multiply equation (ii) by

$13$ we get

$13x -13y= 156$

$13x -y = 48$ [ eq (i) ]

$- \quad + \quad -$
___________________________

$\quad -12 y = 108$

$\Rightarrow y = -9$
Substituting

$y = -9$ in equation (i) we get

$13x -(-9) = 48$

$\Rightarrow 13x = 39$

$\Rightarrow x = 3$

$\therefore$ solution is

$x = 3$ and

$y = -9$

For solving each pair of equation, in this exercise, use the method of elimination by equating coefficients:

$\frac{1}{5}(x-2)=\frac{1}{4}(1-y)$
$26x+3y=-4$

$\frac {1}{5}(x -2) = \frac {1}{4}(1 -y) \Rightarrow 4x +5y = 13 ....(1)$

$26 +3y = -4 ..........(2)$
Multiplying equation no (1) by

$3$ and (2) by

$5$

$12x +15y = 39 ............(3)$

$130x +15y = -20$

$\quad - \quad - \quad$
___________________________

$\quad -115x = 59$

$x = - \frac { 59}{118}$

$x = - \frac {1}{2}$
from (1)

$4 ( - \frac {1}{2} ) +5y = 13$

$5y = 13 +2$

$y = 3$

For solving each pair of equation, in this exercise, use the method of elimination by equating coefficients:

$3-(x-5)=4+2$
$2(x+y)=4-3y$

$3-(x-5)=4+2$

$2(x+y)=4-3y$

$\Rightarrow -x-y = -6$

$\Rightarrow x +y = -6$

$\Rightarrow x +y = 6 ......(1)$

$2x +5y = 4 ............(2)$
Multiplying equation no (1) by (2)

$2x +2y = 12$

$2x + 5y = 4$

$- \quad - \quad -$
________________________

$-3y = 8 \Rightarrow y = \frac {-8}{3}$
From (1)

$x - \frac {8}{3} = 6 \Rightarrow x = \frac {26}{3}$

Solve each of the following pairs of equation by reducing then to a pair of linear equations.
$\dfrac{10}{x+y}+\dfrac{2}{x-y}=4,\dfrac{15}{x+y}-\dfrac{5}{x-y}=-2$

$\dfrac{10}{x+y}+\dfrac{2}{x-y}=4,\dfrac{15}{x+y}-\dfrac{5}{x-y}=-2$

$\dfrac{1}{x+y}=a$

$and \dfrac{1}{x-y}=b,$

$then$

$10a+2b=4.........(1)$

$15a-5b-2......(2)$

$Equation (2)\times 2,gives$

$30a-10b=-4.....(4)$

$Equation(3)-(4),gives$

$16b=16\Rightarrow b=\dfrac{16}{16}=1$

$Substitute b = 1 in equation (1),gives$

$10a +2 (1)=4\Rightarrow 10a+2=4\Rightarrow 10a=4-2=2\Rightarrow a=\dfrac{2}{10}=\dfrac{1}{5}$

$But a = \dfrac{1}{x+y}\Rightarrow \dfrac{1}{5}=\dfrac{1}{x+y}\Rightarrow x+y=5......(5)$

$b=\frac{1}{x-y}\Rightarrow 1=\dfrac{1}{x-y}\Rightarrow x-y=1....(6)$

$Equation (5)+(6), gives$

$2x=6\Rightarrow x=\dfrac{6}{2}=3$

$Subtitute x = 3 in equation (5),gives$

$3+y=5\Rightarrow y=5-3=2$

$Solution(x,y)=(3,2)$

For solving each pair of equation, in this exercise, use the method of elimination by equating coefficients:

$\frac{5y}{2}-\frac{x}{3}=8$
$\frac{y}{2}+\frac{5x}{3}=12$

The given pair of linear equations are

$\frac {5y}{2} -\frac {x}{3} = 8$

$\Rightarrow - \frac {x}{3} +\frac {5y}{2} = 8 ............(i) [ on simplifying ]$

$\frac {y}{2} + \frac {5x}{3} = 12$

$\Rightarrow \frac {5x}{3} + \frac {y}{2} = 12 ..............(ii)$ [ On simpliflying ]
Multiplying equation (i) by

$5$ we get

$- \frac {5x}{3} + \frac {25 y}{2} = 40$

$\frac { 5x}{3} +\frac {y}{2} = 12$ .....[equation (ii) ]

$\quad + \quad + \quad +$
_______________________________

$\quad \frac { 26 y}{2} = 52$

$\Rightarrow 13 y = 52$

$\Rightarrow y = 4$
Substituting

$y = 4$ in equation (i) we get

$- \frac {x}{3} + \frac { 5 (4)}{2} = 8$

$\Rightarrow -\frac {x}{3} = 8 -10$

$\Rightarrow x =6$

$\therefore$ solution is

$x = 6$ and

$y = 4$

Check whether the following equations are consistent or inconsistent , solve them graphically.
$\dfrac{4}{3} x + 2y = 8 , 2x + 3y = 12$

$\dfrac{4}{3} x + 2y = 8 , 2x + 3y = 12$

$\Rightarrow 4x + 6y = 24 , 2x + 3y = 12$

$\Rightarrow 2x + 3y = 12 , 2x + 3y = 12$

$\dfrac{a_{1}}{a_{2}} = \dfrac{4}{2} = 2$ ,

$\dfrac{b_{1}}{b_{2}} =\dfrac{6}{3} =2$

$\dfrac{c_{1}}{c_{2}}= \dfrac{24}{12}=2$

$\therefore \dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} = \dfrac{c_{1}}{c_{2}}$

$\therefore$ the linear equation are consistent

$\therefore$ They have infinite solutions.

Solve the following simultaneous equation graphically.
$x - 3y = 1$ ; $3x - 2y + 4 =0$

$x - 3y = 1$

$y=\dfrac{x-1}{3}$

$|x=1,4,7|$

$|y=0,1,2|$

$3x - 2y + 4 = 0$

$y=\dfrac{3x+4}{2}$

$|x=0,2,4|$

$|y=2,5,8|$

The solution of the given equations is the point of intersection of the two lines i.e (-2, -1)

1812031_1850813_ans_6802edc19a4f4809b2bf3ac193662852.png

The sum of digits of a two digit number is $11 .$ If the digit at ten's place is increased by $5$ and the digit at unit' place is decreased by $5$ , the digits of the number are found to be reversed.Find the original number.

Let

$x$ be the number at ten's place
and

$y$ be the number at the unit's place.
So the number is

$10x+y$ .
The sum of digit of a two digit number is

$11$ .

$\Rightarrow x+y=11.....(i)$
if the digit at ten's place is increased by

$5$
and the digit at unit place is decreases by

$5$ ,
the digits of the number are found to be reversed.

$\Rightarrow 10(x+5)+(y-5)=10y+x$

$\Rightarrow 9x-9y=-45$

$\Rightarrow x-y=-5.......(ii)$
Subtracting equation (i) from equation (ii)., we get:

$x-y=-5$

$x+y-11\quad \quad \quad [Equation(i)]$

$-\quad -\quad - \quad \quad \quad [subtracting]$
____________

$\quad -2y=-16$

$\Rightarrow y=8$
Subtracting

$y=8$ in equation(i), we get

$x+8=11$

$\Rightarrow x=3$
The number is

$10x+y=10(3)+8=38$

In the figure given above, ABCD is a parallelogram. Find the value of x and y.

Given: ABCD is a parallelogram

Therefore,

AD = BC

AB = DC

Thus,

$4y = 3x - 3$ [since AD = BC ]

$\Rightarrow 3x - 4y = 3$ ........(I)

Also,

$6 y + 3 = 4x$ [ since AB = DC]

$4x - 6y = 2 .....(ii)$

Now multiplying eq. (1) by 4 and eq. (2) by 3 and Subtracting (1) from (2) we get,

$12x-12x -18y+16y=6-12$

$-2y=-6$

$y = 3$

Now putting

$y=3$ in eq. (1) we get,

$3x-4\times3 = 3$

$3x = 15$

$x = 5$

Therefore,

$x = 5, y = 3$

Solve using the method of elimination by equating coefficients:

$2x-3y-3=0$

$\dfrac{2x}{3}+4y+\dfrac{1}{2}=0$

The given equations are

$2x-3y-3=0$

$\dfrac{2x}{3}+4y+\dfrac{1}{2}=0$

$\Rightarrow 4x +24 y +3= 0$

$\Rightarrow 4x +24 y = -3$

$\Rightarrow 2x -3y = 3 ....(1)$

$\Rightarrow 4x +24 y = -3 ..... (2)$

Multiplying equation no (1) by

$8$

Adding,

$16x -24 y = 24 ..........(3)$

$4x +24y = -3$

________________________

$\quad 20x = 21$

$\Rightarrow x = \dfrac {21}{20}$

From (1)

$2 ( \dfrac { 21}{20}) - 3y = 3$

$\Rightarrow -3y = 3 -\dfrac {21}{10}$

$\Rightarrow -3y = \dfrac {9}{10}$

$\Rightarrow y = \dfrac {-3}{10}$

Hence, the solutios are

$\ x = \dfrac {21}{20}$ and

$y = \dfrac {-3}{10}$

From the pair of leaner equation in the following problems , and find their solution graphically .
5 pencils and 7 pens together cost Rs , 50 whereas 7 pencils and 5 pens together cost Rs . 46 .Find the cost of a pencils and on one pen .

Cost of each pencil be

$Rs 'X'$
Cost of pen be

$Rs 'Y'$

$5x+ 7y = 50...........(i)$

$7x + 5y = 46 ...........(ii)$
Equation (i)

$y=\dfrac{50-5x}{7}$

$x=3,0$

$y=5,7.143$

Equation (ii)

$y=\dfrac{46-7x}{5}$

$x=3,5$

$y=5,2.2$

Point intersect

$(3,5)$

$\therefore x=3,y=5$

Cost of each pencil be

$Rs 3$
Cost of pen be

$Rs 5$

1835639_1867295_ans_a0285040443e4b2b8b22d4921b194a5b.png

$A$ and $B$ both have some pencils . if $A$ gives $10$ pencils to $B$ , then $B$ will have twice as many as $A$ . And if $B$ gives $10$ pencils to $A$ , then they will have the same number of pencils . how many pencils does each have ?

Let the number of pencils with

$A=x$
and the number of pencil with

$B=y$ .
If

$A$ gives

$10$ pencils to

$B$ ,

$y+10=2(x-10)$

$2x-y=30.....(1)$
If

$B$ gives to pencils to

$A$

$y-10=x+10$

$x-y=-20 \quad .....(2)$

$2x-y=30\quad .....(1)$

$- \quad +\quad -$
______________

$\quad -x=-50$

$x=50$
From

$(1)$

$2(50)-y=30$

$y=70$
Thus,

$A$ has

$50$ pencils and

$B$ has

$70$ pencils.

The graph of $3x + 2y = 6$ meets the $x=axis$ at point $P$ and the $y-axis$ at point $Q$ . Use the graphical method to find the co-ordinates of points $P$ and $Q$ .

by choosing a couple of values for

$x, -1, 0, 1$

calculate the corresponding

$y$ values.

$x$

$-2$

$0$

$2$

$y$

$6$

$3$

$0$

According to the graph, the line intersect

$x$ axis at

$(2, 0)$ and

$y$ at

$(0, 3)$ .

The coordinates of

$P(x-axis)$ and

$Q(y-axis)$ are

$(2, 0)$ and

$(0, 3)$ respectively.

1805095_1843642_ans_30ae35165c494d1fbd5d609cbfa3700b.png

Solve:
$11(x-5)+10(y-2)+54=0$
$7(2x-1)+9(3y-1)=25$

$11(x-5) + 10(y -2) +54 = 0$ (given )

$\Rightarrow 11x -55 +10 y - 20 +54 = 0$

$\Rightarrow 11x + 10y - 21 = 0$

$\Rightarrow 11 x + 10y = 21 ...........(1)$

$7(2x -1) + 9(3y -1) = 25$ (given)

$\Rightarrow 14x -7 + 27 y -9 = 25$

$\Rightarrow 14x +27y -16 = 25$

$\Rightarrow 14x +27y = 41$ .........(2)
Multiplying equation (2) by

$27$ and equation

$(2)$ by

$10$ we get

$297x + 270 y = 567 .........(3)$

$140 =x + 270 y = 410 ........(4)$
Subtracting eq (4) from eq (3) we get

$157x = 157$

$\Rightarrow x = 1$
Substituting

$x =1$ in equation (1) we get

$11 \times 1 + 10y = 21$

$\Rightarrow 10y = 10$

$\Rightarrow y = 1$

$\therefore$ Solution set is

$x = 1$ and

$y = 1$

Given the linear equation $2x + 3 y - 8 = 0$ write linear equation in two variable such that the geometrical representation of the pair so formed is :
Intersecting lines

Linear equation for intersection lines:
2x + 3y - 8 = 0
3x + 2y - 7 = 0

$a_{1} = 2 , b_{1} = 3 , c_{1} = -8$

$a_{2} = 3 , b_{2} = 2 ,c_{2} = - 7$

$\dfrac{a_{1}}{a_{2}} = \dfrac{2}{3} ,\dfrac{b_{1}}{b_{2}} = \dfrac{3}{2}$
Here, when

$\dfrac{a_{1}}{a_{2}} \neq \dfrac{b_{1}}{b_{2}}$
Geometrical representation in intersection lines.

Solve the following pair of linear equation by the substituting method .
$0 . 2x + 0 . 3y = 1.3$
$0.4 x + 0.5y = 2.3$

$0 . 2 x + 0 . 3 y = 1.3 .......(i)$

$0 . 4 x + 0 . 5y = 2.3 .........(ii)$
From eqn

$(i)$

$0 . 2x + 0 . 3y = 1.3$

$x =\dfrac{ 1.3 - 0 . 3y}{0.2}$

Substitute the value of

$x$ in eq (ii)

$0.4\times \left(\dfrac{1.3-0.3y}{0.2}\right)+0.5y=2.3$

$2\times \left(1.3-0.3y\right)+0.5y=2.3$

$2.6-0.6y+0.5y=2.3$

$-0.1y=-0.3$

$y=3$

Substitute the value of

$y$ in eq (i)

$0 . 2 x + 0 . 3(3) = 1.3$

$0.2x=1.3-0.9$

$0.2x=0.4$

$x=2$

Form the pair of linear equation for the following problems and find their solution by substituting method .
The taxi charges in a city consist of a fixed charge together with the charge for the distance covered . For distance of $10$ the charge paid is Rs $105$ and for a journey of $15$ km the charge paid is Rs $155$ . What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of a $25$ km ?

Let the fixed charge for taxi

$=Rs\ x$

And the variable cost per km

$=Rs\ y$

Total cost =fixed charge +variable charge

Given that for a distance of

$10\ km$ the charge paid is

$Rs\ 105$

$x+10y=105$

$x=105-10y...(i)$

Given that for journey of

$15\ km$ the charge paid is

$Rs155$

$x+15y=155$

Put the value of x we get

$105-10y+15y=155$

$5y=155-105$

$5y=50$

$y=10$

put

$y=10$ in eq (i)

$x=105-10(10)=105-100=5$

People have pay for travelling a distance of 25km

$=x+25y$

$=5+25(10)=5+250=255$

Apeaple have to pay Rs 255 for 25 km

Solve the following pair of linear equation by the substituting method .
$\dfrac{3x}{2} - \dfrac{5y}{3} = -2$
$\dfrac{x}{3} + \dfrac{y}{2} = \dfrac{13}{2}$

$\dfrac{3x}{2} - \dfrac{5y}{3} = - 2 ....(i)$

$\dfrac{x}{3} + \dfrac{y}{2} = \dfrac{13}{6} ..........(ii)$

From eqn (1)

$\dfrac{3x}{2} - \dfrac{5y}{3} = - 2$

$\dfrac{9x-10y}{6}=-2$

$9x=10y-12$

$x=\dfrac{10y-12}{9}....(iii)$

From eqn (ii)

$\dfrac{x}{3} + \dfrac{y}{2} = \dfrac{13}{6}$

$\dfrac{2x+3y}{6}=\dfrac{13}{6}$

$2x+3y=13......(iv)$

Substitute the value of

$x$ in eqn (iv)

$2\left(\dfrac{10y-12}{9}\right)+3y=13$

$\dfrac{20y-24+27y}{9}=13$

$47y-24=117$

$47y=117+24$

$47y=141$

$y=3$

$x=\dfrac{10y-12}{9}....(iii)$

$x=\dfrac{10(3)-12}{9}$

$x=2$

Form the pair of linear equation for the following problems and find their solution by substituting method .
The difference between two number is $26$ and one number is three times the other. Find them.

Let one number be

$'x'$
another number be

$'y'$
Their difference is

$26$

$x - y = 26 ................(i)$
One number is three times the other,

$x = 3y$

Put in eq(i)

$3y-y=26$

$2y=26$

$y=13$

$\therefore x=3\times13=39$

one number is

$'39'$
another number be

$'13'$

Form the pair of linear equation for the following problems and find their solution by substituting method .
The larger of two supplementary angles axceeds the smaller by $18$ degrees . Find them .

$In\space two\space supplementary\space angles\space$

$, let\space larger\space angle \space be 'x' \space and \space smaller \space be "y"$

$The \space property \space of Supplementary \space Angle$

$Sum\space of\space two \space angle\space$

$= 180 ^{o}$

$\implies x + y = 180 .........(i)$

$The\space larger\space of\space two\space supplementary\space angles\space exceeds \space the\space smaller\space by\space$

$18$

$degrees.$

$\implies x=y+18..........(2)$

$Substituting\space eq(2) \space in \space eq(1)$

$y+18+y=180$

$\implies2y=162$

$\implies y=81^{o}$

$\implies x=18+81=109^{o}$

$Angles \space are \space 109^{o} \space , \space 81^{o}$

Solve the following pair of linear equation by the substituting method .
$s - t = 3$
$\dfrac{s}{3} + \dfrac{t}{2} = 6$

$s - t = 3$

$\dfrac{s}{3} + \dfrac{t}{2} = 6$

$s - t = 3 .....(i)$

$\dfrac{s}{3} + \dfrac{t}{2} = 6 .....(ii)$
From eqn (i)

$s - t = 3$

$- t = 3 - s$

$t= - 3 + s$
Substituting the value of 't' in eqn (ii)

$\dfrac{s}{3} + \dfrac{- 3 + s}{2} = 6$

$5 s - 9 = 6 \times 6$

$5 s - 9 = 36$

$5 s = 36 + 9$

$5 s = 45$

$s=9$

Substitute the value of

$s$ in eq (i)

$t=-3+9$

$t=6$

Solve the following pair of linear equation by the substituting method .
$3x - y = 3$
$9x - 3y = 9$

$3x - y = 3 ...(i)$

$9x - 3y = 9 ...(ii)$

From eqation (i)

$y=3x-3$

substitute the value of

$y$ in equation (ii)

$9x - 3y = 9$

$9 x - 3 (3x-3 ) = 9$

$9x + 9 - 9x = 9$

$9=9$
Here we can give any value for 'x' i.e infinite solutions are there

$y = 3x- 3$

Means

$x = 0$ then

$y = -3$

$x = 1 , y = 0$

$x = 2 . y = 3$

Solve the following pair of linear equation by the substituting method .
$x + y = 14$
$x - y = 14$

$x + y = 14 ....(i)$

$x - y = 4 .........(ii)$
In eqn

$(i) x+ y = 14$

$x = 14 - y$
Substituting the value of 'x' in equation (ii) , we have

$x - y = 4$

$(14 - y ) - y =4$

$14 - y - y = 4$

$14 - 2y = 4$

$- 2y = 4 - 14$

$- 2y = -10$

$2y = 10$

$x = 5$
Substituting

$y = 5$ in

$x = 14 - y$ ,

$x = 14 - y$

$= 14 - 5$

$x = 9$

$x = 9 , y = 5$

Form the pair of linear equation for the following problems and find their solution by substituting method .
The coach of a cricket them buys $7$ bats and $6$ balls for Rs $3800$ . Later , she buys $3$ bats and $5$ balls for Rs $1750$ Find the cost of each bat and each ball.

Let the cost of each bat be Rs x.

Cost of each ball be Rs y

$7x + 6y = 3800\rightarrow (1)$

$3x + 5y = 1750$

$3x = 1750 - 5y$

$x=\frac{1750 - 5y}{3}$

Putting in (1)

$\frac{1750 - 5y}{3} + 6y=3800$

$12250-35y+18y=11400$

y=850

$\therefore x=\frac{1750 - 5\times 850}{3}=\frac{-2500}{3}$

-ve value cannot be possible.

As cost cannot be -ve.

Solve the following pair of linear equation by the substituting method .
$\sqrt{2x} + \sqrt{ 3y} = 0$
$\sqrt{3x} - \sqrt{8 y} = 0$

$\sqrt{2}x + \sqrt{3}y = 0 .....(i)$

$\sqrt{3}x - \sqrt{8}y = 0 ......(ii)$

From eqn (i)

$\sqrt{2}x + \sqrt{3}y = 0$

$\sqrt{2}x = -\sqrt{3}y$

$x=-\dfrac{\sqrt{3}}{\sqrt{2}}y$

Put in eqn (ii)

$\sqrt{3}x - \sqrt{ 8} y = 0$

$\sqrt{3} \left ( -\dfrac{\sqrt{3}}{\sqrt{2}} y \right ) - \sqrt{8}y = 0$

$-\dfrac{3}{\sqrt{2}}y-2\sqrt{2}y=0$

$\dfrac{-3y-4y}{\sqrt{2}}=0$

$-7y=0$

$y=0$

$x=-\dfrac{\sqrt{3}}{\sqrt{2}}y$

$x=-\dfrac{\sqrt{3}}{\sqrt{2}}(0)$

$x=0$

Form the pair of linear equation for the following problems and find their solution by substituting method .
A fraction becomes $\dfrac{9}{11}$ if $2$ is added to both the numerator and the denominator . If $3$ is added to both the numerator and the denominator it becomes $\dfrac{5}{6}$ .Find the fraction .

If the Numerator is 'x' and denominator is 'y' , then the Fraction is

$\dfrac{x}{y}$
Adding 2 to both Numerator and denominator , then fraction

$\dfrac{9}{11}$

$11 ( x + 2) = 9 ( y + 2 )$

$11 x + 22 = 9y + 18$

$11 x - 9 y + 22 - 18= 0$

$11 x - 9y + 4 = 0...(i)$
Is 3 is added to both numerator and denominator , then ther fraction is

$\dfrac{5}{6}$

$6(x+3)=5(y+3)$

$6x+18=5y+15$

$x=\dfrac{5y-3}{6}...(ii)$

substitute the value of

$x$ in eqn (i)

$=11\times \dfrac{5y-3}{6} - 9y + 4 = 0$

$=\dfrac{55y-33}{6}-9y+4=0$

$=\dfrac{55y-33-54y+24}{6}=0$

$=y=9$

Substitute the value of

$y$ in eq (ii)

$x=\dfrac{5(9)-3}{6}$

$x=7$

$\therefore$ fraction is

$\dfrac{7}{9}$

Write the number of solutions of the following pair of liner equations

$x + 2y - 8 = 0 , 2x + 4y = 16$

$\dfrac{a_{1}}{a_{2}} = \dfrac{1}{2} , \dfrac{b_{1}}{b_{2}} = \dfrac{2}{4} = \dfrac{1}{2} , \dfrac{c_{1}}{c_{2}} = \dfrac{- 8}{- 16} = \dfrac{1}{2}$

$\therefore \dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} = \dfrac{c_{1}}{c_{2}}$

The pair of linear equations has infinite many solution

Solve the following pair of linear equations graphically and write nature of solution.
$2x - y = 4; x + y = -1$

Given linear pair

$2x - y = 4$

$2x - y - 4 = 0$ ........(i)

$x + y = -1$

$x + y + 1 = 0$ .......(ii)
Comparing equation (i) and (ii) by linear pair

$a_{1}x + b_{1}y + c_{1} = 0$ and

$a_{2}x + b_{2}y + c_{2} = 0$

$a_{1} = 1, b_{1} = -1$ and

$c_{1} = 4$

$a_{2} = 1, b_{2} = 1$ and

$c_{1} = -1$

$\dfrac{a_1}{a_2} = \dfrac{2}{1}, \dfrac{b_1}{b_2} = \dfrac{-1}{1}$
Thus

$\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$
Linear pair is consistent which all will have unique solutions.
Graphical Method:
By equation (i),

$2x - y = 4$
Putting

$x = 0, 2 \times 0 - y = 4$

$y = -4$
Putting

$x = 1, 2 \times1 - y = 4$

$-y = 4 - 2=2$

$y=-2$
Putting

$x = 2, 2 \times 2 - y = 4$

$4 - y = 4$

$y = 0$
Table

$: 1$

$x=0,1,2$

$y=$
By equation (ii),

$x + y = -1$
Putting

$x = 0. 0 + y = -1$

$y = -1$
Putting

$x = 1, +1 + y = -1$

$y = -1 - 1$

$y = -2$
Putting

$x = 2,$

$2 + y = -1$

$y = -3$
Table

$: 2$

$x=0,1,2$

$y=-1,-2,-3$
By plotting the points of Table and 2 we get two straight lines.
From above graph, it is clear that both the straight lines cut each other at point

$P(1, -2).$
Thus,

$x = 1, y = -2$ are required solution

1869441_1875448_ans_a9ecc6c5b32c43b4b814b4d911d532d4.png

Solve the following pair of linear equations graphically and find the coordinates of the triangle so formed with the y-axis and the lines.
$4x - 5y = 20,$
$3x + 5y = 15$

Given, pair of linear equation

$4x - 5y = 20$ ..........(i)
and

$3x + 5y = 15$ .........(ii)
From equation (i),

$4x - 5y = 20$

$5y = 4x - 20$

$y = \dfrac{4x - 20}{5}$
Putting

$x = 0,$

$y = \dfrac{4 \times 0 - 20}{5} = \dfrac{-20}{5}$

$y = -4$
Putting

$x = 2, y = \dfrac{4 \times 2 - 20}{5} = \dfrac{8 - 20}{5}$

$y = \dfrac{-12}{5}$

$y = -2.4$
Putting

$x = 5, y = \dfrac{4 \times 5 - 20}{5} = \dfrac{20 - 20}{5}$

$y = 0$
Table : 1

$x=0,\,\,\,2,\,\,\,5$

$y=-4,-2.4,0$
From equation (ii),.

$3x + 5y = 15$

$5y = 15 - 3x$

$y = \dfrac{15 - 3x}{5}$
Putting

$x = 0, y = \dfrac{15 - 3 \times 0}{5}$

$= \dfrac{15 - 0}{5}$

$y = 3$
Putting

$x = 5, y = \dfrac{15 - 3 \times (5)}{5}$

$= \dfrac{15 - 15}{5}$

$y = 0$
Putting

$x = -5, y = \dfrac{15 - 3 \times (-5)}{5}$

$= \dfrac{15 + 15}{5}$

$= 6$
Table : 2

$x=0,5,-5$

$y=3,0,6$
Plot the points obtained from Table (1) and (2) on graph paper. By joining these points two straight lines are obtained.
From the given graph it is clear that two lines intersect each other at point

$P(5, 0)$

$\therefore x = 5$ and

$y = 0$ are required solution.

$(0, 3), (0, -4)$ and

$(5, 0)$ are co-ordinates of vertices of

$\Delta ABP$ formed by two straight lines at y-axis.

1869461_1875472_ans_a615ad59cea5412ba8e3ec40990bb48e.png

Solve the following pair of linear equations, graphically and find the coordinates of that points where lines represented by these cuts y-axis.
$2x - 5y + 4 = 0; 2x + y - 8 = 0$

Given pair of linear equations

$2x - 5y + 4 = 0$ ..........(i)
and

$2x + y - 8 = 0$ ......(ii)
From equation (i),

$2x - 5y + 4 = 0$

$2x = 5y - 4$

$x = \dfrac{5y - 4}{2}$
Putting

$y = 0, x = \dfrac{5 \times 0 - 4}{2} = \dfrac{-4}{2} = -2$
Putting

$y = 2, x = \dfrac{5 \times 2 - 4}{2} = \dfrac{6}{2} = 3$
Putting

$y = -2, x = \dfrac{5 \times (-2) - 4}{2} = \dfrac{-14}{2} = -7$
Table : 1

$x=-2,3,7$

$y=0,2,-2$
By equation (ii),

$2x + y - 8 = 0$
or

$y - 2 + 8 = 0$
Putting

$x = 4, y = -2 \times 4 + 8$

$= -8 + 8 = 0$
Putting

$x = 3, y = -2 \times 3 + 8$

$= -6 + 8$

$= 2$
Putting

$x = 2, y = -2 \times 2 + 8$

$= -4 + 8$

$= 4$
Table : 2

$x=4,3,2$

$y=0,2,4$

Plot the points from Table (1) and (2) on graph.
By joining these points two straight lines are obtained.
From given graph it is clear that two straight lines intersect each other at point

$P(3, 2).$

$\therefore$ Its required solutions are

$x = 3$ and

$y = 2.$
and two straight lines cuts the y-axis at

$(0, 0.8)$ and

$(0, 8).$

1869451_1875470_ans_5b770afd9ad442c5ab9c1b50935f8056.png

Solve the following pair of linear equations graphically and find the coordinates of that points where lines are represented by these cuts y-axis.
$3x + 2 = 12; 5x - 2y = 4$

Given pair of linear equations

$3x + 2y = 12$ .........(i)
and

$5x - 2y = 4$ ...........(ii)
From equation (i),

$3x + 2y = 12$

$2y = 12 - 3x$

$y = \dfrac{12 - 3x}{2}$
Putting

$x = 0, y = \dfrac{12 - 3 \times 0}{2} = \dfrac{12}{2} = 6$

Putting

$x = 2, y = \dfrac{12 - 3 \times 2}{2} = \dfrac{6}{2} = 3$

Putting

$y = 4, y = \dfrac{12 - 3 \times 4}{2} = \dfrac{0}{2} = 0$
Table : 1

$x=0,2,4$

$y=6,3,0$

From equation (ii),

$5x - 2y = 4$

$2y = 5x - 4$

$y = \dfrac{5x - 4}{2}$
Putting

$x = 2, y = \dfrac{5 \times 2 - 4}{2} = \dfrac{6}{2} = 3$

Putting

$x = 0, y = \dfrac{5 \times 0 - 4}{2} = \dfrac{-4}{2} = -2$

Putting

$x = 4, y = \dfrac{5 \times 4 - 4}{2} = \dfrac{16}{2} = 8$
Table : 2

$x=2,0,4$

$y=3,-2,8$
Plot the points from Tables (1) and (2) on graph paper. By joining these points two straight lines are obtained.
From the above graph, it is clear that two straight lines intersect each other at point

$P(2, 3).$

$\therefore x = 2$ and

$y = 3$ are required solutions and two straight lines cut y-axis at

$(0, 6)$ and

$(0, -2).$

1869455_1875471_ans_5a4d16186207437898717aab8b96f4f5.png

Draw the graphs of the equations 5x - y = 5 and 3x - y = 3 . Determine the coordinates of the vertices of the triangle formed by these lines and the y -axis .

vertices are A(-3,0),B(-5,0),C(1,0)
1959840_1867778_ans_2dfed3ad8e58404ebcc8c99e9b8e799f.png

Roohl travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60km by bus and if she travels 100 km by bus , it takes 10 minutes longer . Find the speed of the train and the bus separately

Let speed of the train be x km / hr . and speed of the bus is y km / hr

$\dfrac{240}{x} + \dfrac{60}{y} = 4$

$\dfrac{200}{x} + \dfrac{100}{y} = \dfrac{25}{6}$

Let,

$u=\frac{1}{x},v=\frac{1}{y}$

$240u + 60v = 4\rightarrow (1)$

$200u + 100v = \dfrac{25}{6}\rightarrow (2)$

$Subtracting(2)\times 6from(1)\times 10$

$2400u-1200u=40-25$

$u=\frac{1}{80}$

$\therefore v=\frac{1}{60}$

$\therefore x=80,y=60$

Answer

Solve the following systems of simultaneous linear equations by the method of elimination by equating the coefficient.
$\dfrac{x}{7} + \dfrac{y}{3} = 5$
$\dfrac{x}{2} - \dfrac{y}{9} = 6$

The given equations are

$\dfrac{x}{7} + \dfrac{y}{3} = 5$ .........(i)
and

$\dfrac{x}{2} - \dfrac{y}{9} = 6$ ..........(ii)
Multiplying equation (i) by

$\dfrac{1}{3}$

$\dfrac{x}{21}+\dfrac{y}{9}=\dfrac{5}{3}$

and then adding in equation (ii), we get

$\dfrac{x}{21} + \dfrac{x}{2} = \dfrac{5}{3} + 6$

$\Rightarrow \dfrac{2x + 21x}{42} = \dfrac{5 + 18}{3}$

$\Rightarrow \dfrac{23x}{42} = \dfrac{23}{3}$

$\Rightarrow x = \dfrac{42}{3} = 14$
Putting

$x = 14$ in (i), we get

$\dfrac{14}{7} + \dfrac{y}{3} = 5$

$\Rightarrow \dfrac{y}{3} = 5 - 2 \Rightarrow y = 9$
Hence the required solution is

$x = 14$ and

$y = 9$

Solve graphically:
$3x - 4y = 1; -2x + \dfrac{8}{3}y = 5$

From the given equation

$3x - 4y = 1$

$\Rightarrow 4y = 3x - 1$

$\Rightarrow y = \dfrac{3x - 1}{4}$
(Expressing y in terms of x) .......(i)
when

$x = 3, y = \dfrac{3 \times 3 - 1}{4} = \dfrac{8}{4} = 2$
when

$x = 0, y = \dfrac{3 \times 0 - 1}{4} = -0.25$
when

$x = -1, y = \dfrac{3 \times (-1) - 1}{4}$

$= \dfrac{-3 - 1}{4} = -1$
Table for equation (i)

$x=$	$3,$	$0,$	$-1$
$y=$	$2,$	$-0.25,$	$-1$

Similarly from equation

$-2x + \dfrac{8}{3}y = 5$

$\Rightarrow -6x + 8y = 15$

$\Rightarrow 8y = 15 + 6x$

$\Rightarrow y = \dfrac{15 + 6x}{8}$ ..........(i)
when

$x = 0, y = \dfrac{15 + 6 \times 0}{8} = \dfrac{15}{8} = 1.8$
when

$x = -2.5, y =\dfrac{15+6\times(-2.5)}{8}$

$=\dfrac{15-15}{8}=0$

$\Rightarrow x = \dfrac{-15}{6}$

$\Rightarrow x = -2.5$
Table for equation (ii)

$x=$	$0,$	$-2$ . $5$
$y=$	$1.8,$	$0$

Now plotting and joining the points from table (i) and (ii) on the same graph paper we find that these lines are parallel to each other it means no solution.
Hence, the system of equations has no solutions.

1867387_1878701_ans_8519b816c9e9479ebb65c64d10b2fb01.png

Solve the following systems of simultaneous linear equations by the method of elimination by equating the coefficient.
$2x + y = 13$
$5x - 3y = 16$

The given equations are

$2x + y = 13$ .......(i)
and

$5x - 3y = 16$ .......(ii)
Multiplying equation (i) by

$3$

$6x+3y=39$

and then adding in (ii),
we get

$11x = 55$

$x = \dfrac{55}{11} = 5$
Substituting the value of x in (i), we get

$2 \times 5 + y = 13$

$\Rightarrow 10 + y = 13$

$\Rightarrow y = 13 - 10 = 3$
Hence the required solution is

$x = 5, y = 3.$

Solve the following systems of simultaneous linear equations by the method of substitution.
$8x + 5y = 9$
$3x + 2y = 4$

The given equations are

$8x + 5y = 9$ .......(i)
and

$3x + 2y = 4$ ........(ii)
From (ii)

$2y = 4 - 3x$

$\Rightarrow y = \dfrac{4 - 3x}{2}$ .......(iii)
Substituting the value of

$y = \dfrac{4 - 3x}{2}$ in (i),
we get

$8x + 5 \left ( \dfrac{4 - 3x}{2} \right ) = 9$

$\Rightarrow 8x + \dfrac{20 - 15x}{2} = 9$

$\Rightarrow 16x + 20 - 15x = 18$

$\Rightarrow x = 18 - 20$

$\Rightarrow x = -2$
Now putting

$x = -2$ in (iii), we get

$y = \dfrac{4 - 3 \times (-2)}{2} = \dfrac{4 + 6}{2}$

$\Rightarrow y = \dfrac{10}{2} = 5$
Hence, the required solution is

$x = -2$ and

$y = 5.$

Solve the following systems of simultaneous linear equations by the method of substitution.
$3x + 2y = 11$
$2x + 3y = 4$

The given equations are

$3x + 27 = 11$ .........(i)
and

$2x + 3y = 4$ .......(ii)
From (i)

$2y = 11 - 3x$

$\Rightarrow y = \dfrac{11 - 3x}{2}$ ..........(iii)
Substituting the value of y from (iii) in (ii),
we get

$2x + 3 \left ( \dfrac{11 - 3x}{2} \right ) = 4$

$\Rightarrow 4x + 33 - 9x = 8$

$\Rightarrow -5x = -25$

$\Rightarrow x = 5$
Putting

$x = 5$ in (iii), we get

$y = \dfrac{11 - 3 \times 5}{2} = \dfrac{11 - 15}{2}$

$= \dfrac{-4}{2} = -2$
Hence the required solution is

$x = 5$ and

$y = -2.$

Solve the following systems of simultaneous linear equations by the method of substitution.
$4x - 5y = 39$
$2x - 7y = 51$

The given equations are

$4x - 5y = 39$ ........(i)
and

$2x - 7y = 51$ .........(ii)
From equation (ii),

$2x - 51 = 7y$

$\Rightarrow y = \dfrac{2x - 51}{7}$ .......(iii)
Substituting the value of y in (i), we get

$4x - 5 \left ( \dfrac{2x - 51}{7} \right ) = 39$
or

$28x - 10x + 255 = 273$
or

$18x = 18$
or

$x = \dfrac{18}{18} = 1$
Putting

$x = 1$ in (iii), we get

$y = \dfrac{2 \times 1 - 51}{7}$

$\Rightarrow y = \dfrac{-49}{7} = -7$
Hence the required solution is

$x = 1, y = -7.$

Solve the following systems of simultaneous linear equations by the method of substitution.
$x + 2y = -1$
$2x - 3y = 12$

The given equations are

$x + 2y = -1$ .......(i)
and

$2x - 3y = 12$ ........(ii)
From (i)

$x = -1 - 2y$ ......(iii)
Substituting this value of x in (ii), we get

$2(-1 - 2y) - 3y = 2$

$\Rightarrow -2 - 4y - 3y = 12$

$\Rightarrow -7y = 14$

$\Rightarrow y = -2$
Putting

$y = -2$ in (iii), we get

$x = -1 - 2 \times (-2) = -1 + 4 = 3$
Hence the required solution is

$x = 3$ and

$y = -2.$

Solve the following systems of simultaneous linear equations by the method of elimination by equating the coefficient.
$0.4x + 0.3y = 1.7$
$0.7x - 0.2y = 0.8$

The given equations are

$0.4x + 0.3y = 1.7$ ..........(i)

$0.7x - 0.2y = 0.8$ .........(ii)
Multiplying equation (i) by

$2$ and (ii) by

$3$

$0.8+0.6=3.4$

$2.1-0.6=2.4$

and then adding, we get

$2.9x = 5.8$

$\Rightarrow x = 2$
Substituting the value of x in (i), we get

$0.4 \times 2 + 0.3y = 1.7$

$\Rightarrow 0.3y = 1.7 - 0.8$

$\Rightarrow 0.3y = 0.9$

$\Rightarrow y = 3$

Solve the following systems of simultaneous linear equations by the method of substitution.
$5x - 2y = 19$
$3x + y = 18$

The given equations are

$5x - 2y = 19$ ......(i)
and

$3x + y = 18$ .....(ii)
From (ii),

$y = 18 - 3x$ .......(iii)
Substituting from (iii) in (i), we get

$5x - 2(18 - 3x) = 19$

$\Rightarrow 5x - 36 + 6x = 19$

$\Rightarrow 11x = 55$

$\Rightarrow x = 5$
Putting

$x = 5$ in (iii), we get

$y = 18 - 3 \times 5$

$y = 18 - 15 = 3$
Hence the required solution is

$x = 5$ and

$y = 3.$

Solve the following systems of simultaneous linear equations by the method of elimination by equating the coefficient.
$11x + 15y = -23$
$7x - 2y = 20$

The given equations are

$11x + 15y = -23$ .........(i)
and

$7x - 2y = 20$ ......(ii)
Multiplying equation (i) by

$2$ and (i) by

$15$

$22x+30y=-46$

$105x-30y=300$

and then adding, we get

$127x = 254$

$\Rightarrow x = 2$
Putting

$x = 2$ in (ii), we get

$7 \times 2 - 2y = 20$

$\Rightarrow 14 - 20 = 2y$

$\Rightarrow 2y = -6$

$\Rightarrow y = -3$
Hence the required solution is

$x = 2$ and

$y = -3.$

Solve the following systems of simultaneous linear equations by the method of elimination by equating the coefficient.
$\dfrac{1}{2x} - \dfrac{1}{y} = -1$
$\dfrac{1}{x} + \dfrac{1}{2y} = 8$

The given equations are

$\dfrac{1}{2x} - \dfrac{1}{y} = -1$ .........(i)
and

$\dfrac{1}{x} + \dfrac{1}{2y} = 8$ .............(ii)
Multiplying equation (i) by

$\dfrac{1}{2}$

$\dfrac{1}{4x}-\dfrac{1}{2y}=-\dfrac{1}{2}$

and then adding, we get

$\dfrac{1}{4x} + \dfrac{1}{x} = \dfrac{-1}{2} + \dfrac{8}{1}$

$\Rightarrow \dfrac{1 + 4}{4x} = \dfrac{-1 + 16}{2}$

$\Rightarrow \dfrac{5}{4x} = \dfrac{15}{2} \Rightarrow 60x = 10$

$\Rightarrow x = \dfrac{1}{6}$
Putting

$x = \dfrac{1}{6}$ in (ii), we get

$\dfrac{1}{\dfrac{1}{6}} + \dfrac{1}{2y} = 8$

$\Rightarrow 6 + \dfrac{1}{2y} = 8$

$\Rightarrow \dfrac{1}{2y} = 2 \Rightarrow y = \dfrac{1}{4}$
Hence the required solution is

$x = \dfrac{1}{6}$ and

$y = \dfrac{1}{4}$

Solve the following systems of simultaneous linear equations by the method of elimination by equating the coefficient.
$\dfrac{5}{x + y} - \dfrac{2}{x - y} = -1$
$\dfrac{15}{x + y} + \dfrac{7}{x - y} = 10$

The given equations are

$\dfrac{5}{x + y} - \dfrac{2}{x - y} = -1$ ...........(i)

$\dfrac{15}{x + y} + \dfrac{7}{x - y} = 10$ .....(ii)
Let

$\dfrac{1}{x + y} = a, \dfrac{1}{x - y} = b$

$\Rightarrow 5a - 2b = -1$ .........(iii)

$15a + 7b = 10$ ...........(iv)
Multiplying (iii) by

$7$ and (iv) by

$2$

$35a-14b=-7$

$30a+14b=20$

and then adding, we get

$65a = 13$

$\Rightarrow a = \dfrac{13}{65} = \dfrac{1}{5}$
Putting

$a = \dfrac{1}{5}$ in (iii), we get

$5 \times \dfrac{1}{5} - 2b = -1$

$\Rightarrow 2 = 2b \Rightarrow b = 1$
Now

$\dfrac{1}{x + y} = \dfrac{1}{5}$

$\Rightarrow x + y = 5$ .........(v)

$\dfrac{1}{x - y} = 1$

$\Rightarrow x - y = 1$ .......(vi)
Adding (v) and (vi), we get

$2x = 6$

$\Rightarrow x = 3$ and

$3x + y = 5$

$\Rightarrow y = 2$
Hence the required solution is

$x = 3$ and

$y = 2.$

Solve the following systems of simultaneous linear equations by the method of elimination by equating the coefficient.
$3x - 7y + 10 = 0$
$y - 2x = 3$

The given equations are

$3x - 7y + 10 = 0$ ..........(i)
and

$y - 2x = 3$ .........(ii)
Multiplying equation (i) by

$2$ and (ii) by

$3$

$6x-14y=-20$

$3y-6x=9$

and then adding, we get

$-11y = -11$

$\Rightarrow y = 1$
Putting

$y = 1$ in (ii), we get

$1 - 2x = 3 \Rightarrow 1 - 3 = 2x$

$\Rightarrow 2x = -2$

$\Rightarrow x = -1$
Hence the required solution is

$x = -1$ and

$y = 1$

Solve the following system of equations
$2x + 3y = 9$
$3x + 4y = 5$

The given equations are

$2x + 3y = 9$ .........(i)
and

$3x + 4y = 5$ ......(ii)
Multiplying equation (i) by

$3$ and (ii) by

$2$

$6x+9y=27$

$6x+8y=10$

and then subtracting, we get

$y = 17$
Putting the value of

$y = 17$ in equation (i), we get

$2x + 3 \times 17 = 9$

$\Rightarrow 2x + 51 = 9$

$\Rightarrow 2x = 9 - 15$

$\Rightarrow 2x = -42$

$\Rightarrow x = -21$
Hence the required solution is

$x = -21$ and

$y = 17.$

Solve the following systems of simultaneous linear equations by the method of elimination by equating the coefficient.
$x + 2y = \dfrac{3}{2}$
$2x + y = \dfrac{3}{2}$

The given equations are

$x + 2y = \dfrac{3}{2}$ .........(i)
and

$2x + y = \dfrac{3}{2}$ ..........(ii)
Multiplying equation (ii) by

$2$

$4x+2y=3$

and then subtracting from (i), we get

$-3x = -\dfrac{3}{2}$

$\Rightarrow x = \dfrac{1}{2}$
Putting the value of x in (i), we get

$\dfrac{1}{2} + 2y = \dfrac{3}{2}$

$2y = \dfrac{3}{2} - \dfrac{1}{2}$

$\Rightarrow 2y = \dfrac{2}{2} = 1$

$\Rightarrow y = \dfrac{1}{2}$
Hence the required solution is

$x = \dfrac{1}{2},$

$y = \dfrac{1}{2}.$

A two digit number is $4$ times its sum of digits and $2$ times the product of its digits. Find the number.

Let a two digit number be

$10x + y,$ where

$y$ is unit place digit and

$x$ is tens place digit

$10x + y = 4(x + y)$

$\Rightarrow 10x + y = 4x + 4y$

$\Rightarrow 6x - 3y = 0$

$\Rightarrow 2x = y$ .......(i)
and

$10x + y = 2xy$ ....(ii)
Substituting

$y = 2x$ in (ii), we get

$10x + 2x = 2x \times 2x$

$\Rightarrow 4x^{2} - 12x = 0$

$\Rightarrow 4x(x - 3) = 0$

$\Rightarrow x = 0, 3$
where

$x = 0, y = 0$
when

$x = 3, y = 2 \times 3 = 6$

$\therefore$ Required number is

$10x + y = 10 \times 3 + 6 = 36$

Solve the question :
Thilo and Toby buy some boats and trains from the toy shop.
The cost of one boat is b cents and the cost of one train is t cents.
(i) Toby buys $3$ boats and $4$ trains for $$5.70$ .
Complete this equation.
$3b + 4t$ = ..................
(ii) Thilo buys $1$ boat and $2$ trains for $$2.40$ .
Write this information as an equation.
................... = ......................
(iii) Solve your two equations to find the cost of a boat and the cost of a train.
You must show all your working.

1952102_1912837_ans_e3d7b298e3694576945736a15c56ce16.jpg

The difference between two numbers is $26$ and the larger number exceeds thrice of the smaller number by $4$ . Find the numbers.

Let the larger number be

$y$ and the smaller number be

$x$ .
According to question,

$y-x=26$ ....(1)
and

$y=3x+4$ .....(2)
Substituting the value of

$y$ from equation (2) in equation (1), we get

$3x+4-x=26$
or

$2x=26-4$
or

$2x=22$
or

$x=11$
Putting this value of

$x$ in equation (1), we get

$y-11=26$
or

$y=26+11=37$
Hence, the numbers are

$11$ and

$37$

Solve the following systems of simultaneous linear equations by the method of elimination by equating the coefficient.
$8v - 3u = 5uv$
$6v - 5u = -2uv$

The given equations are

$8v - 3u = 5uv$ .......(i)
and

$6v - 5u = -2uv$ .........(ii)
Dividing equation (i) and (ii) both side by uv, we get

$\dfrac{8}{u} - \dfrac{3}{v} = 5$ .........(iii)

$\dfrac{6}{u} - \dfrac{5}{v} = -2$ .......(iv)
Multiplying equation (iii) by 5 and equation (iv) by 3

$\dfrac{40}{u}-\dfrac{15}{v}=25$

$\dfrac{18}{u}-\dfrac{15}{v}=-6$

and then subtracting, we get

$\dfrac{22}{u} = 31$

$\Rightarrow u = \dfrac{22}{31}$
Putting the value of

$u = \dfrac{22}{31}$ in (iii) we get

$\dfrac{8 \times 31}{22} - \dfrac{3}{v} = 5$

$\Rightarrow \dfrac{124}{11} - \dfrac{3}{v} = 5 \Rightarrow \dfrac{124}{11} - 5 = \dfrac{3}{v}$

$\Rightarrow \dfrac{124 - 55}{11} = \dfrac{3}{v} \Rightarrow \dfrac{69}{11} = \dfrac{3}{v}$

$\Rightarrow v = \dfrac{23}{11}$

Hence the required solution is

$u = \dfrac{22}{31}$ and

$v = \dfrac{23}{11}.$

Solve for 'u' and 'v', $2(3u-v)=5uv$ ; and $2(u+3v)=5$ uv.

$\begin{matrix} 2\left( { 3x-v } \right) =5uv \\ \Rightarrow 2\left( { \frac { { 3x } }{ { uv } } -\frac { v }{ { uv } } } \right) =\frac { { 5uv } }{ { uv } } \\ 2\left( { \frac { 3 }{ v } -\frac { 1 }{ u } } \right) =5\to (i) \\ and\, \, 2\left( { u+3v } \right) =5uv \\ \Rightarrow 2\left( { \frac { u }{ { uv } } +\frac { { 3v } }{ { uv } } } \right) =\frac { { 5uv } }{ { uv } } \\ \Rightarrow 2\left( { \frac { 1 }{ v } +\frac { 3 }{ u } } \right) =5\to (ii) \\ Let\, \, \, \, \, \frac { 1 }{ v } =x,\, \, \frac { 1 }{ u } =y \\ from\, \, equation\, \, (i)\, and\, \, (ii) \\ \Rightarrow 2\left( { 3x-y } \right) =5 \\ \Rightarrow 3x-y=\frac { 5 }{ 2 } \to (ii) \\ and\, \, \\ \Rightarrow 2\left( { x+3y } \right) \, =5\, \\ \Rightarrow x+y=\frac { 5 }{ 2 } \, \\ { { adding } }\, \, { { equation(iii) } }\, \, { { and } }\, \, { { (iv) } } \\ { { 3x-y= } }\frac { { { 5 } } }{ { { 2 } } } { { \times 1 } } \\ x+3y=\frac { 5 }{ 2 } \times 3 \\ 3x-y=5/2 \\ \underline { 3x } \underline { + } 9y=\frac { { 15 } }{ { -2 } } \\ -10y=-5, \\ y=\frac { 5 }{ { 10 } } =\frac { 1 }{ 2 } =\frac { 1 }{ u } \\ u=2 \\ Putting\, \, =\frac { 1 }{ 2 } in\, \, equation\, \, \left( { iii } \right) \\ \Rightarrow 3x\frac { { -1 } }{ 2 } =\frac { 5 }{ 2 } \\ \Rightarrow 3x=3 \\ x=1=\frac { 1 }{ v } \\ \therefore v=1\, \, \, and\, \, u=2\, \, \, Ans. \\ \end{matrix}$

Determine the point on the graph of the linear equation $2x+5y=19$ , whose ordinate is $1\dfrac{1}{2}$ times its abscissa.

Let the abscissa be

$a$

Then ordinate

$=\cfrac{3}{2a}$

New coordinates of point is

$(a,\cfrac{3}{2a})$

Given that is a point on line

$2x+5y=19$

now

$2(a)+5(\cfrac{3}{2a})=19$

$2a+\cfrac{15a}{2}=19$

$4a+15a=38$

$19a=38$

$a=2$

$\therefore$ abscissa

$=2$

ordinate

$=\cfrac{3(2)}{2}=3$

$\therefore$ the required point

$(2,3)$

Solve the following simultaneous equations
$\dfrac {1}{2}x-y=\dfrac {5}{2};x+\dfrac {1}{3}y=\dfrac {8}{3}$

$\dfrac{1}{2}x-y=\dfrac{5}{2}$ .......

$(1)$

or,

$\dfrac{x-2y}{2}=\dfrac{5}{2}\Rightarrow x-2y=5$ .........

$(2)$

$x+\dfrac{1}{3}y=\dfrac{8}{3}$ ...........

$(3)$

or,

$\dfrac{3a+y}{3}=\dfrac{8}{3}\Rightarrow 3x+y=8$ ..........

$(4)$

Solving

$(2)$ and

$(4)$

$3x-6y=15$

$3x+y=8$

_____________

$-7y=7$

$\therefore y=-1$

Putting

$y=-1$ in

$(2)$

$x-2(-1)=5$

or,

$x=5/2=3$

$\therefore x=3$

$x=3, y=-1$

Solve the following pairs of equations by reducing them to a pair of linear equations:
(i) $\dfrac{1}{2x}+\dfrac{1}{3y}=2 ; \dfrac{1}{3x}+\dfrac{1}{2y}=\dfrac{13}{6}$

(ii) $\dfrac{2}{\sqrt{x}}+\dfrac{3}{\sqrt{y}}=2 ; \dfrac{4}{\sqrt{x}}-\dfrac{9}{\sqrt{y}}=-1$

(iii) $\dfrac{4}{x}+3y=14 ; \dfrac{3}{x}-4y=23$

(iv) $\dfrac{5}{x-1}+\dfrac{1}{y-2}=2 ; \dfrac{6}{x-1}-\dfrac{3}{y-2}=1$

(v) $\dfrac{7x-2y}{xy}=5 ; \dfrac{8x+7y}{xy}=15$

(vi) $6x + 3y = 6xy ; 2x + 4y = 5xy$

(vii) $\dfrac{10}{x+y}+\dfrac{2}{x-y}=4 ; \dfrac{15}{x+y}-\dfrac{5}{x-y}=-2$

(viii) $\dfrac{1}{3x+y}+\dfrac{1}{3x-y}=\dfrac{3}{4} ; \dfrac{1}{2(3x+y)}-\dfrac{1}{2(3x-y)}=\dfrac{-1}{8}$

(i)

$\dfrac1{2x} + \dfrac1{3y} = 2$

$\dfrac1{3x} + \dfrac1{2y} = \dfrac{13}6$

Lets assume

$\dfrac1x = p$ and

$\dfrac1y = q$ , both equations will become

$\dfrac p2 + \dfrac q3 = 2$

$\Rightarrow 3p + 2q = 12$ ... (A)

$\dfrac p3 + \dfrac q2 = \dfrac{13}6$

$\Rightarrow 2p + 3q = 13$ ...(B)

Multiply (A) by

$3$ and (B) by

$2$ , we get

$9p + 6q = 36$

$4p + 6q = 26$

Subtracting these both, we get

$5p = 10$

$\Rightarrow p = 2$

$\Rightarrow x = \dfrac1p = \dfrac12$

$\therefore q = \dfrac{26 - 8}6 = 3$

$\Rightarrow y = \dfrac1q = \dfrac13$

(ii)

$\dfrac2{\sqrt x} + \dfrac3{\sqrt y} = 2$

$\dfrac4{\sqrt x} - \dfrac9{\sqrt y} = -1$

Lets assume

$\dfrac1{\sqrt x} = p$ and

$\dfrac1{\sqrt y} = q$ , both equations will become

$2p + 3q = 2$ ...(A)

$4p - 9q = -1$ ...(B)

Multiply (A) by

$3$ , we get

$6p + 9q = 6$

$4p - 9q = -1$

Adding these both, we get

$10p = 5$

$\Rightarrow p = \dfrac12$

$\Rightarrow \ \ x = \dfrac1{p^2} = 4$

$\therefore q = \dfrac13$

$\Rightarrow y = \dfrac1{q^2} = 9$

(iii)

$\dfrac4x + 3y = 14$

$\ \ \dfrac3x - 4y = 23$

Lets assume

$\dfrac1x = p$ , both equations become

$4p + 3y = 14$ ...(A)

$3p - 4y = 23$ ...(B)

Multiply (A) by

$4$ and (B) by

$3$ , we get

$16p + 12y = 56$

$9p - 12y = 69$

Adding these both, we get

$25p = 125$

$\Rightarrow p = 5$

$\Rightarrow x = \dfrac1p = \dfrac15$

$\Rightarrow y = -2$

(iv)

$\dfrac5{x-1} + \dfrac1{y-2} = 2$

$\dfrac6{x-1} - \dfrac3{y-2} = 1$

Lets assume

$\dfrac1{x-1} = p$ and

$\dfrac1{y-2}=q$ , both equations become

$5p+q=2$

$6p-3q=1$

Multiply (A) by

$3$ , we get

$15p + 3q = 6$

$6p - 3q = 1$

Adding both of these, we get

$21p = 7$

$\Rightarrow p = \dfrac13$

$\Rightarrow \ \ \ x = 4$

$\therefore q = \dfrac 13$

$\Rightarrow \ \ \ y = 5$

(v)

$\dfrac{7x-2y}{xy} = 5$

$\Rightarrow \dfrac7y - \dfrac2x = 5$

$\dfrac{8x+7y}{xy} = 15$

$\Rightarrow \ \ \dfrac8y + \dfrac7x = 15$

Let

$t=\dfrac{1}{x}$ and

$r=\dfrac{1}{y}$ equation will be

$7r-2t=5\;;\;\; 8r+7t=15$

On solving we get

$t=1\;\; r=1$

$\therefore x=1\;and\; y=1$

(vi)
Dividing both side by xy then we have

$\dfrac{6}{y}+ \dfrac{3}{x}=6\;\;;\;\; \dfrac{2}{y}+\dfrac{4}{x}=5$
Let

$t=\dfrac{1}{x}\;and\; r=\dfrac{1}{y}$ equation will be

$6r+3t=6\;;\;\; 2r+4t=5$ on solving we get

$t=1\;\; r=\dfrac{1}{2}$

$\therefore x=1\;and\; y=2$

(vii)
Let

$t=\dfrac{1}{x+y}\;\; and\;\; r=\dfrac{1}{x-y}$ then equation will be

$10t+2r=4\;;\;\; 15t-5r=-2$ on solving we have

$t=\dfrac{1}{5}\;\;and\;\; r=1$

$\therefore x+y=5 \;\;and\;\;x-y=1$
Hence,

$x=4,y=1$

(viii)

Let

$t=\dfrac{1}{3x+y}\;\; and\;\; r=\dfrac{1}{3x-y}$ then equation will be

$t+r=\dfrac{3}{4}\;;\;\; \dfrac{t}{2}-\dfrac{r}{2}= \dfrac{-1}{8}$

On solving we have

$t=\dfrac{1}{4}\;\;and\;\; r=\dfrac{1}{2}$

$\therefore 3x+y=4 \;\;and\;\;3x-y=2$
Hence,

$x=1,y=1$

Mary told her daughter, Seven years ago, I was seven times as old as you were then. Also three years from now, I shall be three times as old as you will be, Find the present age of Mary and her daughter.

Let the age of Mary be $x$ and the age of her daughter be $y$ .

Seven years ago, Mary was $\left( {x - 7} \right)$ years old and her daughter was $\left( {y - 7} \right)$ years old.

Then, according to the question,

$x - 7 = 7\left( {y - 7} \right)$

$x - 7 = 7y - 49$

$x - 7y = - 42$ ....................(1)

Three years from now, Mary will be $\left( {x + 3} \right)$ years old and her daughter will be $\left( {y + 3} \right)$ years old.

According to question,

$x +3 = 3\left( {y + 3} \right)$

$x + 3 = 3y + 9$

$x - 3y = 6$ ........................(2)

From equation (1) and (2),

Subtract equation $(1)$ from $(2)$ ,

$4y=48$ $\rightarrow y=12$

Put the value of $y$ in equation $(1)$

$x=7\times 12-42$ $\rightarrow y=42$

$x = 42$ and $y = 12$

Therefore, the present age of Mary is 42 and that of her daughter is 12.

Solve:
$\dfrac{2}{x}+\dfrac{2}{3y}=\dfrac{1}{16}$ : $\dfrac{3}{x}+\dfrac{2}{y}=0$

$\dfrac {2}{x}+\dfrac {2}{3y}=\dfrac {1}{6}---(1)\quad \dfrac {3}{x}+\dfrac {2}{y}=0---(2)$

Let

$\dfrac {1}{x}=\alpha$ and

$\dfrac {1}{y}=\beta$

$2\alpha +\dfrac {2}{3}\beta =\dfrac {1}{6}$

or,

$6\alpha +2\beta =\dfrac {3}{16}---(3)$

Solving

$(3)$ and

$(4)$

$\dfrac { 6\alpha +2\beta =\dfrac { 3 }{ 16 } \\ 3\alpha +2\beta =0\\ -\quad -\quad - }{ 3\alpha =\dfrac { 3 }{ 16 } \Rightarrow \boxed {\alpha =\dfrac { 1 }{ 16 } } }$

Putting

$\alpha =\dfrac {1}{16}$ in

$(4)$

$\dfrac {3}{16}+2/3=0\Rightarrow 2\beta =-\dfrac {3}{16}\ \Rightarrow \boxed {\beta =-\dfrac {3}{32}}$

Substituting value of

$\alpha$ and

$\beta$ ,

$\boxed {x=16}$ and

$\boxed {y=-\dfrac {32}{3}=-10\ 2/3}$

A two -digit number is obtained by either multiplying the sum of the digits by $8$ and then subtracting $5$ or by multiplying the difference of the digits by $16$ and then adding $3$ . find the number.

2172835_1795747_ans_a8731126853743d186df5407b3e713c1.jpg

From the pair of linear equations in the following problems , and find their solutions ( if they exist ) by the elimination method :
If we add 1 to the numerator and subtract 1 from the denominator , a fraction reduces toIt becomes $\dfrac{1}{2}$ if we only add 1 to the denominator . What is the fraction ?

Let the original fraction is

$\frac { x }{ y }$

1) From the first condition, we get,

$\frac { x+1 }{ y-1 } =1$

$\therefore x+1=y-1$

$\therefore x-y=-2$ (1)

2) From the second condition,

$\frac { x }{ y+1 } =\frac { 1 }{ 2 }$

$\therefore x=\frac { 1 }{ 2 } \left( y+1 \right)$

$\therefore 2x=y+1$

$\therefore 2x-y=1$ (2)

Subtract equation (1) from (2), we get,

$x=3$

Put this value in equation (1), we get,

$3-y=-2$

$\therefore y=5$

From the pair of linear equations in the following problems , and find their solutions ( if they exist ) by the elimination method :
Five years ago , Nuri was thrice as old as Sonu . Ten years later , Nuri will be twice as old as sonu . How old are Nuri and Sonu ?

Let current age of Sonu is

$x$ years and current age of Nuri is

$y$ years.

1) From the first condition, we can write,

$y-5=3\left( x-5 \right)$

$\therefore y-5=3x-15$

$3x-y=10$ (1)

2) From the second condition, we can write,

$y+10=2\left( x+10 \right)$

$\therefore y+10=2x+20$

$\therefore 2x-y=-10$ (2)

Subtract equation (2) from (1), we get,

$x=20$

Put this value in equation (1), we get,

$3\left( 20 \right) -y=10$

$\therefore 60-y=10$

$\therefore y=50$

Thus, current age of Sonu is 20 years and current age of Nuri is 50 years

State whether True or False.
There is no common point to the graph of the equation: $y = 2x + 3$ and $y = -2x + 3$ .

Given equations:

$y = 2x + 3$ (green line)

and

$y = -2x + 3$ (blue line)

On putting the value of

$x = 0$ in both the equations, we get

$y = 3$ and

$y = 3$ .

So, at

$x = 0$ , both the equations will meet at

$y = 3$ on the graph.
Hence, the given statement is false.

Find the value of $x + y$ , if
$12 x + 13 y = 29$
and $13x + 12y = 21$

Given:

$12x+13y = 29$

$13x+12y = 21$

Adding both the equations, we get

$25x+ 25y = 50$

$\Rightarrow 25(x+y) = 50$

Divide both sides by

$25$ , we get

$x+y = 2$

Verify if the following simultaneous equations are consistent or inconsistent.
I. $2x+5y=17$ , $5x+3y=14$
II. $2x+6y=7$ , $6x+18y=10$

(i) Comparing

$2x+5y=17$ with

$a_1x+b_1y+c_1=0$ , we get,

$a_1=2,b_1=5,c_1=-17$

and

$5x+3y=14$ with

$a_2x+b_2y+c_2=0$ , we get,

$a_2=5,b_2=3,c_2=-14$

Here,

$\dfrac {a_1}{a_2}=\dfrac{2}{5}$ ,

$\dfrac{b_1}{b_2} = \dfrac{5}{3}$

$\because \dfrac{2}{5} \neq \dfrac{5}{3}$

The equations are consistent.

(ii) Comparing

$2x+6y=7$ with

$a_1x+b_1y+c_1=0$ ,we get,

$a_1=2,b_1=6,c_1= -7$ `

and

$6x+18y = 10$ with

$a_2x+b_2y+c_2 =0$ ,we get,

$a_2=6, b_2=18,c_2= -10$

Here,

$\dfrac {a_1}{a_2}=\dfrac{2}{6}$ ,

$\dfrac{b_1}{b_2}=\dfrac{6}{18}$

$\dfrac{c_1}{c_2} =\dfrac{-7}{-10}$

$\because \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq \dfrac{c_1}{c_2}$

$\therefore$ the equations are inconsistent.

If $x + y = 5$ and $x = 3$ , then the value of $y$ .

Given:

$x+y = 5$ , and

$x= 3$

Substituting the value of

$x$ in the given equation, we get

$3+y = 5$

$\Rightarrow y=2$

Find the value of $k$ for which the given simultaneous equations have infinitely many solutions: $4x+y=7, 16x+ky=28$ .

$4x+y=7$ and

$16x+ky=28$

Comparing the above equation with the equation

$a_1x+b_1y=c_1$ and

$a_2x+b_2y=c_2$ respectively.

$a_1=4, b_1=1, c_1=7$ ,

$a_2=16, b_2=k, c_2=28$

Apply the condition for infinitely many solution.

$\displaystyle\frac{a_1}{a_2}=\frac{4}{16}=\frac{1}{4}$ ,

$\displaystyle\frac{b_1}{b_2}=\frac{1}{k}$ and

$\displaystyle\frac{c_1}{c_2}=\frac{1}{4}$

$\therefore$ The condition for the simultaneous equations to have infinitely many solutions is.

$\displaystyle\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

$\therefore \displaystyle\frac{1}{4}=\frac{1}{k}=\frac{1}{4}$

$\therefore \displaystyle\frac{1}{4}=\frac{1}{k}$

$\therefore k=4$ .

Hence the simultaneous equations

$4x+y=7$ and

$16x+ky=28$ will have infinitely many solutions for

$k=4$ .

Solve the following simultaneous equations using graphical method
$x + y = 7$
$x - y = 5$

Clearly, the graphs cut at

$x=6$ and

$y=1$ .
Therefore, solution is

$x=6$ and

$y=1$

Solve the pair of equations by Elimination method.
$2x + y - 5 = 0, \quad 3x - 2y - 4 = 0$

Given:

$2x+y-5=0$

$..........(1)$

$3x-2y-4=0$

$..........(2)$

On multiplying equation

$(1)$ by

$2$ and adding with equation

$(2)$ , we get

$4x+2y-10=0$

$\dfrac{3x-2y-4=0}{7x-14=0}$

$7x=14$

$x=2$

Now putting value of

$x=2$ in eqn.

$(1)$ , we get

or,

$2\times2+y-5=0$

or,

$4+y-5=0$

or,

$y-1=0$

or,

$y=1$

So,

$x=2$ and

$y=1$

Solve the following equations using graphical method.
$x + y = 6$
$x - y = 4$

Given: (1)

$x + y = 6$

$\therefore y = 6 -x$

$x$ =	$1$	$2$	$3$
$y$ =	$5$	$4$	$3$
$(x, y)$	= $(1, 5)$	$(2, 4)$	$(3, 3)$

(2)

$x - y = 4$

$\therefore y = x - 4$

$x$ =	$1$	$2$	$3$
$y$ =	$-3$	$-2$	$-1$
$(x, y)$ =	$(1, -3)$	$(2, -2)$	$(3, -1)$

From the graph, it can be seen clearly that

$(x,y) = (5,1)$ is the solution to the pair of simultaneous equations

Solve the following system of equations.
$7x - 2y = 1$
$3x + 4y = 15$

Given:

$7x - 2y = 1$ ...(1)

$3x + 4y = 15$ .... (2)

First multiply the first equation by

$2$ :

We got,

$14x- 4y = 2$

Now add the first and second equation:

$14x - 4y +3x + 4y = 15 +2$

$17x =17$

so,

$x =1$

By putting

$x$ in any equation we got;

$y = 3$

Solve graphically the equations $2x-y=1$ ; $x+2y=8$

We find three points for each equation, by choosing three values of

$x$ and computing the corresponding

$y$ values. We'll put our results in tables.
Line 1:

$2x-y=1$

$y=2x-1$
Substituting

$x=-1, 0, 1$ in the above equation, we find

$y=2x-1$

$x$	$-1$	$0$	$1$
$2x$	$-2$	$0$	$2$
$y=2x-1$	$-3$	$-1$	$1$

Line 2:

$x+2y=8$

$2y=-x+8 \Rightarrow y=-\dfrac{x}{2}+4$
Substituting

$x=-2, 0, 2$ in above equation, we get

$y=-\dfrac{x}{2}+4$

$x$	$-2$	$0$	$2$
$-\dfrac{x}{2}$	$1$	$0$	$-1$
$y=-\dfrac{x}{2}+4$	$5$	$4$	$3$

We plot the points

$\left( -1,-3 \right)$ ,

$\left( 0,-1 \right)$ and

$\left( 1,1 \right)$ in a graph sheet and draw the line through them. Next, we plot the points

$\left( -2,5 \right)$ ,

$\left( 0,4 \right)$ and

$\left( 2,3 \right)$ in the same graph sheet and draw the line through them. We find that the two graphs are intersecting at the point

$\left( 2,3 \right)$ . Hence, the system of equations has only one solution (unique solution|) and the solution is

$x=2$ ,

$y=3$ . Therefore the solution is

$\left( 2,3 \right)$ .
1089099_618913_ans_2922e70484834caebf0924fc5c4615a0.jpg

Solve the following systems of equations graphically:
$3x+y+1=0$
$2x-3y+8=0$

Given: equations of two lines:

$3x+y+1=0$ and

$2x-3y+8 =0$

(1) Intercept of line

$3x+y+1=0$

$x = 0 , y = -1$

$y=0, x = \dfrac{-1}{3}$

Now plot the first line on the graph using intercepts.

(2) Intercepts of the line

$2x-3y+8 = 0$

$x=0, y=\dfrac{8}{3}$

$y = 0, x = -4$

Now plot the second line on the graph using intercepts.

We can see that these two lines intersect a point

$x=-1,y=2$

Solve the following systems of equations graphically:
$x+y=6$
$x-y=2$

We can see that two lines intersect at a point

$x=4,y=2$

Solve the following systems of equations graphically:
$2x+y-3=0$
$2x-3y-7=0$

We can see that two lines intersect at a point

$x=2,y=-1$

Solve the following systems of equations graphically:
$x+y=4$
$2x-3y=3$

We can see that two lines intersect at a point

$x=3,\,y=1$

Show graphically that each one of the following systems of equations has infinitely many solutions:
$x-2y+11=0$
$3x-6y+33=0$

Both lines have same x-intercept and same y-intercept

x-intercept =-11

y-intercept =

$\dfrac {11}{2}$

And both lines have same Slope

Slope =

$\dfrac {-1}{-2}=\dfrac {1}{2}$

$\Rightarrow$ both lines coincide and have infinite solutions

Show graphically that each one of the following systems of equations has infinitely many solutions:
$3x+y=8$
$6x+2y=16$

x	1	3
y	5	-1

Graph of the given equations:

Both lines have same x-intercept and same y-intercept

x-intercept =

$\dfrac{8}{3}$

y-intercept =6

And both lines have same Slope

Slope =

$\dfrac {-3}{1}=\dfrac {-6}{2}$

$\Rightarrow$ both lines coincide and therefore they have infinite intersection points

$\therefore$ this system has infinite solutions

Find the value of a if the three equations are consistent
$(a + 1)^3 x + (a + 2)^3 y = (a + 3)^3$
$(a + 1) x + (a + 2) y = a + 3$
$x + y = 1$ .

Since the given equations are consistent,

$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$ ......(a)

Considering the equations,

$(a+1)^3x+(a+2)^3y=(a+3)^3$ ......(1)

$(a+1)x+(a+2)y=(a+3)$ .....(2)

From (a)

$\dfrac{(a+1)^3}{(a+1)}=\dfrac{(a+2)^3}{(a+2)}=\dfrac{(a+3)^3}{(a+3)}$

Considering 1st and 2nd terms, we have,

$\dfrac{(a+1)^3}{(a+1)}=\dfrac{(a+2)^3}{(a+2)}$

${(a+1)^2}=(a+2)^2$

$a^2+2a+1=a^2+4a+4$

$-3=2a$

$\Rightarrow a=\dfrac{-3}{2}$

Solve the following systems of equations graphically:
$2x-3y+13=0$
$3x-2y+12=0$

We can see that two lines intersect at a point

$x=-2,\,y=3$

Show graphically that each one of the following systems of equations has infinitely many solutions:
$x-2y=5$
$3x-6y=15$

Both lines coincide each other and has infinite solutions

Both lines same slopes, same x-intercept and same y-intercept

Solve the following systems of equations graphically:
$2x+3y=4$
$x-y+3=0$

We can see that two lines intersect at a point

$x=-1,\,y =2$

Determine, by drawing graphs, whether the following system of linear equation has a unique solution or not:
$2x-3y=6,\ x+y=1$

We can see in the graph that two lines at a point

$\left (\dfrac {9}{5},\dfrac {-4}{5}\right)$

$\therefore$ System of equations has a unique solution

For what value of $k$ the following system of equations will be inconsistent?
$4x+6y=11$
$2x+ky=7$

For the system to be inconsistent, lines must be parallel.

$\Rightarrow$ ratio of respective

$'x'$ and

$'y'$ coefficients must be equal

$\Rightarrow \dfrac {4}{2}=\dfrac {k} {6}$

$\Rightarrow k=12$

the following system of linear equations,
$2x-y-4=0$ and $x+y+1=0$ .

Given equations are

$2x-y-4=0---eqn(1)$

$x+y+1=0---eqn(2)$

Adding

$eqn(1)$ and

$eqn(2)$ , we get

$3x-3=0$

$x=1$ .

Substituting value of

$x$ in

$eqn(2)$ we get

$y=-2$

So value

$(x-y)=1-(-2)=3$ .

Solve each of the following pairs of equations by the elimination method.
$8x+5y=9$
$3x+2y=4$

$8x+5y=4...........(i)$

$93x+2y=4..........(ii)$

Multiply

$(i)$ by

$2$ and

$(ii)$ by

$5$

$16x+10y=8$

$465x+10y=20$

Subtract both the equations

$449x=12$

$x=\dfrac{12}{449}$

$y=\dfrac{340}{449}$

Find the value of $(x+2y)$ satisfying the equations,
$x+y=3$ and $3x-2y=4$ .

Given equations are

$x+y=3---eqn(1)$

$3x-2y=4---eqn(2)$

Multiplying

$eqn(1)$ with

$2$ we get:

$2x+2y=6---eqn(3)$

Adding

$eqn(3)$ and

$eqn(2)$ gives us value

$x=2$

Substituting

$x=2$ in

$eqn(1)$ , we get

$y=1$

So we get

$(x+2y)=2+(2 \times 1) = 4$

Solve for $x$ and $y$ : $47x+31y=63,\ 31x+47y=15$

$47x + 31y = 63$ ---(i)

$31x + 47y = 15$ ---(ii)

multiplying (i) by

$31$ and multiplying (ii) by

$47$ we get

$1457x + 961y = 1953$ ---(iii)

$1457x + 2209y = 705$ ---(iv)

subtracting (iv) from (iii) we get

$-1248y=1248$

$\Rightarrow y = - 1$

putting

$y=-1$ in (i)

$47x + 31 \times - 1 = 63$

$\Rightarrow 47x = 94$

$\Rightarrow x = 2$

Therefore

$x=2$ and

$y=-1$

Solve:
$a= b + 2$
$2a - b =7$
and hence find out $2a+b$ .

$a=b+2$ ---(1)

$2a=b+7$ ----(2)

(2) - (1) gives,

$2a-a=b+7-b-2$

$a=5$

Substitute for

$a$ in (1),

$5=b+2$

$b=3$

To find

$2a+b$

$=2(5)+3$

$=13$

Solve each of the following pairs of equations by the elimination method.
$2x+3y=8$
$4x+6y=7$

$2x+3y=8......(i)$

$4x+6y=7......(ii)$

Multiply

$(i)$ by

$4$

$4x+6y=16......(iii)$

LHS of

$(ii)$ and

$(iii)$ are same but RHS is different

Hence the solution does not exist.

For each real value of $c$ , the pair of equations $x-2y=8;5x-10y=c$ has a unique solution. Justify whether it is True or False.

$x - 2y = 8$

$5x - 10y = c$

Here

$\cfrac{{{a_1}}}{{{a_2}}} = \cfrac{1}{5},\cfrac{{{b_1}}}{{{b_2}}} = \cfrac{{ - 2}}{{ - 10}} = \cfrac{1}{5},\cfrac{{{c_1}}}{{{c_2}}} = \cfrac{8}{c}$

Since

$\cfrac{{{a_1}}}{{{a_2}}} = \cfrac{{{b_1}}}{{{b_2}}}$

Therefore it cannot have unique solution for any value of

$c$

Hence, The given statement is false.

Meena went to a bank to withdraw Rs.She asked the cashier to give her Rs. 50 and Rs. 100 notes only. Meena got 25 notes in all. Find how many notes of Rs. 50 and Rs. 100 she received ?

Solve: $0.2p + 0.3q = 1.3$
$0.4p + 0.5q = 2.3$

Given,

$0.2p + 0.3q = 1.3$ ......(1) and

$0.4p + 0.5q = 2.3$ ......(2).

Now, (2)

$-2\times$ (1) gives,

$-0.1q=-0.3$

or,

$q=3$ .

Using value of

$q$ in (1) we get,

$p=2$ .

Solve the following pair of linear equations by the substitution method.
$x + y = 14$
$x - y = 4$

Given,

$x + y = 14$

or,

$x=14-y$ .......(1)

$x - y = 4$ ........(2).

Now substituting the value of

$x$ from (1) in (2) we get,

$14-2y=4$

or,

$2y=10$

or,

$y=5$ .

Now substituting the value of

$y$ in (1) we get,

$x=9$ .

$x=9,y=5$ .

Solve :3x + 4y = 25, 5x - 6y = -9 by elimination method.

$3x + 4y = 25$

$...\left( 1 \right)$

$5x - 6y = - 9$

$...\left( 2 \right)$

Multiply (1) by 3 and (2) by 2

$9x + 12y = 75$

$\left( + \right)$ ${10x - 12y}$ ${ = - 18}$

$19x = 57$

$\boxed{x = 3}$

Put x = 3 in (2)

$5x - 6y = - 9$

$5\left( 3 \right) - 6y = - 9$

$\,6y = 15 + 9$

$6y = 24$

$\boxed{y = 4}$

By the method of substitution, find the solution set of the pair of linear equations $x+2y=5$ and $3.x+7y=17$

let

$x + 2y = 5 \to x=5-2y$ ------ (1)

$3x+7y=17$ ------ (2)

From (1) put in (2)

$y=2$

put

$y=2$ in (1)

$x=1$

solve:
$4x + \frac{{x - y}}{8} = 17$
$2y + x - \frac{{5y + 2}}{3} = 2$

$4x+\dfrac{x-y}{8}=17$

$33x-y=136$ this is equation 1

$2y+x-\dfrac{5y+2}{3}$

$3x+y=8$ this is second equation

by solving we get

$x=4,y=-4$

Solve for $x$ and $y$ :
$11x + 15y + 23 = 0,7x - 2y - 20 = 0$

$11x + 15y + 23 = 0........(1)$

$7x - 2y - 20 = 0........(2)$

After multiplying equation

$(1)$ by

$2$ and equation

$(2)$ by

$15$ on addition we get,

$\begin{aligned}{}2\left( {11x + 15y + 23} \right) + 15\left( {7x - 2y - 20} \right) &= 0\\22x + 30y + 46 + 105x - 30y - 300& = 0\\127x - 254 &= 0\\127x &= 254\\x &= 2\end{aligned}$

Substitute

$x=2$ in equation

$(1),$

$\begin{aligned}{}22 + 15y + 23 &= 0\\15y + 45 &= 0\\15y &= - 45\\y &= - 3\end{aligned}$

Solve: $x-y=0;2x-y=2$ .

Solution:-

$x - y = 0 ........... \left( 1 \right)$

$2x - y = 2 .......... \left( 2 \right)$

On subtracting

${eq}^{n} \left( 1 \right)$ from

$\left( 2 \right)$ , we have

$2x - y - \left( x - y \right) = 2 - 0$

$2x - y - x + y = 2$

$x = 2$

Substituting the value of x in

${eq}^{n} \left( 1 \right)$ , we have

$2 - y = 0$

$\Rightarrow \; y = 2$

Hence,

$x = 2$ and

$y = 2$ .

Solve the equation by substitution method
$2\,x\, + \,3\,y\, = \,9\,,\,\,3\,x\, + \,4\,y\, = \,5$

$2x + 3y = 9$ (1)

$3x + 4y = 5$ (2)

By (1) $2x = 9 - 3y$

$x = {{9 - 3y} \over 2}$

Putting in (2) , we get

$\displaystyle 3\left( {{{9 - 3y} \over 2}} \right) + 4y = 5$

$27 - 9y + 8y = 10$

$17 = y$

$y = 17$

$\displaystyle x = {{9 - 51} \over 2}$

$\displaystyle = {{ - 42} \over 2}$

$x = - 21$

Solve $x-y+2=0$
$7x+9y=130$

$x - y + 2 = 0$ ------- (1)

$7x + 9y = 130$ ------(2)

multiply equation 1 by

$7$ .

we get ,

$7x-7y+14=0$ ---- (3)

Now subtract (3) from (2).

we get ,

$y=9$

put

$y=9$ in (1)

we get ,

$x=7$

Solve the following pairs of linear equations :
$2x + 3y = 2xy, \, 6x + 12y = 7 xy$

$2x+3y=2xy$ ............ (1)

$6x+12y=7xy$ ........... (2)

Multiply the equation (1) by

$3$ and subtract the result from equation (2),

$6x+12y=7xy$ ........... (2)

$6x+9y=6xy$ ............ (3)

$6x+12y-6x-9y=7xy-6xy$

$3y=xy$

$3y-xy=0$

$y(3-x)=0$

$y=0$ or

$x=3$

Substitute

$y=0$ in equation (1),

$2x+0=0$

$x=0$

So, one solution is

$x=0, y=0$

Substitute

$x=3$ in equation (1),

$2(3)+3y=2(3)y$

$6+3y=6y$

$3y=6$

$y=2$

So, the second solution is

$x=3, y=2$

Solve the following equations.
(i) $10 - q = 6$ (ii) $2t - 5 = 3$

Given the equations,

(i)

$10 - q = 6$

or,

$q=10-6$

or,

$q=4$ .

(ii)

$2t - 5 = 3$

or,

$2t=8$

or,

$t=4$ .

The sum of two fraction is $\dfrac{5}{8}$ . One fraction is greater by $1$ that the other fraction. What are the two fraction.

Let one fraction

$= x$

Let other fraction

$= y$

According to problem

$x+y=\dfrac{5}{8}$ ......(i)

$x+1=y$ ......(ii)

Put equation (ii) in equation (i)

$x+x+1=\dfrac{5}{8}$

$2x=\dfrac{5}{8}-1=\dfrac{-3}{8}$

$x=\dfrac{-3}{16}$

Put value of

$x$ in equation (ii)

$y=\dfrac{13}{16}$

$\therefore$ One fraction

$=\dfrac{-3}{16}$ , other fraction

$=\dfrac{13}{16}$

$a+b=7$ and $a-b=3$ find $a$ and $b$

Consider the given equations,

$a+b=7$ ……….. (1)

$a-b=3$ …………. (2)

From equation (2),

$a=3+b$

Put, the value of $a$ in equation (1), we get

$b+3+b=7$

$2b=4$

$b=2$

Now, put the value of $b$ in equation (2), we get

$a-2=3$

$a=5$

Hence, $a=5,\,b=2$ . This is the answer.

Solve the following equation
$x+3y=6 \\ 2x-3y=12$

Consider the given equation.

$x+3y=6$ …….. (1)

$2x-3y=12$ ……… (2)

On adding both equation (1) and (2), we get

$3x=18$

$x=6$

Now, put the value of $x$ in equation (1), we get

$6+3y=6$

$3y=6-6$

$3y=0$

$y=0$

Hence, the value of $x$ is $6$ and $y$ is $0$ .

Solve the following pair of linear equations :

$\begin{gathered} px + qy = p - q \hfill \\ qx - py = p + q \hfill \\ \end{gathered}$

$px+qy=p-q$ ......(1)

Multiplying by

$p$ on both sides,

$p^2x+pqy=p^2-pq$ ......(2)

$qx-py=p+q$ ......(3)

Multiplying by

$q$ on both sides,

$q^2x-pqy=pq+q^2$ ......(4)

On adding equation (2) and (4), we get

$p^2x+pqy+q^2x-pqy=p^2-pq+pq+q^2$

$(p^2+q^2)x=p^2+q^2$

$x=1$

Substituting the value of

$x$ in equation (1), we get-

$p\times 1+qy=p-q$

$qy=-q$

$y=-1.$

Hence,

$x=1,y=-1$ is the final solution of the given equation.

Solve by elimination method

$x+y=5$
$2x-3y=4$

$x+y=5$

$-(1)$

$2x-3y=4$

$-(2)$

Multiplying eqn

$(1)$ with 3 and adding to eqn

$(2)-$

$\ \ \ \ \ \ 3x+3y=15$

$+\ \ \ 2x-3y=4$

__________________

$\ \ \ \ \ \ \ 5x+0y=19$

$\therefore x = \dfrac{19}{5} = 3.8$

Substituting in eqn

$(1)-$

$y = 5- x = 5-3.8=1.2$

$\therefore x=3.8,y=1.2$

Solve : $2x + 5y + 72 = 30$
$3x - 6y + 52 = 15$

$2x+5y=-42...(1)$

$3x-6y=-37...(2)$

$(1)\times 3 - (2)\times 2$

$6x+15y=-126$ ...(3)

$6x-12y=-74$ ...(4)

Sloving (3) and (4), we get

$y=\dfrac{-52}{27}$

Substituting the value of y in eqn (1), we get

$x=\dfrac{-437}{27}$

Solve $3x+4y=10$ and $2x-2y=2$ by substitution method.

By substitution method,

$3x+4y=10$ &

$2x-2y=2\Rightarrow 2(x-y)=2\Rightarrow x-y=1$ ___ (2)

$\therefore x=y+1\rightarrow (1)$ .

Substituting in equation (2).

$\therefore 3x+4y=10\Rightarrow 3(y+1)+4y=10$

$\Rightarrow 3y+3+4y=10\Rightarrow 7y=7\Rightarrow y=1$

$\therefore x=1+1=2$

$\therefore x=2, y=1$

$2x + 3y = 8\\2x = 2 + 3y$ Find the values of $x$ and $y$

$2x+3y=8$ ---------- ( 1 ) [ Given ]

$2x=2+3y$ --------- ( 2 ) [ Given ]

$\therefore$

$x=\dfrac{2+3y}{2}$ ----- ( 3 )

Now, substituting value of

$x$ in equation ( 1 ),

$2\left(\dfrac{2+3y}{2}\right)+3y=8$

$2+3y+3y=8$

$2+6y=8$

$6y=8-2$

$6y=6$

$\therefore$

$y=1$

Now substituting value of

$y=1$ in equation ( 3 ),

$x=\dfrac{2+3(1)}{2}$

$x=\dfrac{5}{2}$

$\therefore$

$x=2.5$

$\therefore$ The values of

$x$ and

$y$ are

$2.5,\,1$ .

Solve the following pair of equation by the elimination method.
$x+y=14,x-y=4$

$x+y=14$ .....(1)

$x-y=4$ ......(2)

On adding 1 and 2, we get,

$2x=18$

$x=9$

Putting this value in equation 2, we get,

$9-y=4$

$y=5$

Hence,

$x=9$ and

$y=5$ .

Solve the equation by substitution method $2x + 3y = 9$ , $3x + 4y = 5$

We have,

$2x+3y=9$ ……. (1)

$3x+4y=5$ …… (2)

Now,

$y=\dfrac{9-2x}{3}$

On putting the value of $y$ , in equation (2), we get

$3x+4\left( \dfrac{9-2x}{3} \right)=5$

$3x+\dfrac{36-8x}{3}=5$

$9x+36-8x=15$

$x=15-36$

Now, put the value of $x$ in equation (1), we get

$y=\dfrac{9-2\left( -21 \right)}{3}$

$y=\dfrac{9+42}{3}$

$y=\dfrac{51}{3}=17$

Hence, this is the answer.

Solve by elimination method :
$2x\ +\ 5y\ =\ 12$
$7x\ +\ 3y\ =\ 13$

$2x+5y=12........(1)$

$7x+3y=13.........(2)$

Multiply equation

$(1)$ with

$3$ and multiply equation

$(2)$ with

$5$

$6x+15y=36$

$35x+15y=65$

$-29x=-29$

$x=1$

$x=1$ value substitute with equation

$(1)$

$2(1)+5y=12$

$2+5y=12$

$y=2$

Solve each pair of equation by using the substitution method.
$x+3y=-1$
$2x-3y=12$

$x+3y=-1---(i)$

$2x-3y=12---(ii)$

From equation

$(i)$ , we get

$\Rightarrow \ x=-1-3---(iii)$

substituting value of

$x$ in equation

$(iii)$ we get

$2(-1-3y)-3y=12$

$\Rightarrow \ -2-6-3y=12$

$\Rightarrow \ 9y=14$

$\Rightarrow \ y=-\dfrac {14}{9}\ \Rightarrow x=-1+\dfrac {42}{9}$

$=\dfrac {11}{3}$

Solve the following pair of linear (simultaneous) equations by the method of elimination:
$2x+3y=8$
$4x+6y=7$

$2x+3y=8\_\_\_(i)$

$4x+6y=7\_\_\_(ii)$

eq.

$(i)$ and eq.

$(ii)$ are parallel

$\therefore$ infinite solution are possible.

Solve for $x$ and $y$
$6 x - 9 y = 15$
$6 x - 8 y = 14$

$\\6x-9y=15\>\>\rightarrow(1)\\6x-8y=14\>\>\rightarrow(2)\\subtracting\>eq^n\>(1)\>and\>eq^n\>(2),\>we\>get\\-y=1\\\therefore\>y=-1\\then\\6x-9y=15\\or\>x=(\dfrac{15+9y}{6})\\=(\dfrac{15-9}{6})=(\frac{6}{6})=1\\\therefore\>x=1$

Solve the following system of equations by substitution method:
$x+y=2 \\ x-y=2$

$x+y = 2 ...(1)$

$x-y = 2 ...(2)$

$x = 2+y$

$\therefore 2+y+y = 2$

$\therefore 2+2y = 2$

$\therefore 2y = 0$

$\therefore y = 0$

$\therefore x = 2$

$\therefore (x,y) = (2,0)$

Solve the following equations :
$x+2y=2,\ 2x+3y=3$

$x+2y=2$ ......(1)

$2x+3y=3$ ........(2)

In equation 1,

$x=2-2y$

Therefore, in equation 2,

$2(2-2y)+3y=3$

$4-4y+3y=3$

$y=1$

Therefore,

$x=0$

Hence,

$x=0$ and

$y=1$ .

Solve the following system of equations by substitution method:
$2x+3y=11 \\ 2x-4y=-24$

Given: Two linear equations in two variables.

$2x+3y = 11 ...(1)$

$2x-4y = -24 ...(2)$

Using eq.(1) we get,

$2x = 11-3y$

$x = \dfrac{11-3y}{2}$ ...........(3)

Putting this value of

$x$ in (2) we get,

$2\times \dfrac{(11-3y)}{2}-4y = -24$

$11+24 = 7y$

$\therefore y = \dfrac{35}{7} = 5$

Putting this value of

$y$ in (3) we get,

$\therefore x = \dfrac{11-3(5)}{2} = \dfrac{-4}{2}$

$\therefore x = -2$

$\therefore (x,y) = (-2,5)$

1121223_1165018_ans_7a95f8b6b66e45c38cff07c5bba13b4e.jpeg

Solve : $px + qy = p - q$
$qx - py = p + q$

$px+qy=p-q$ ............. (1)

$qx-py=p+q$ .............(2)

Multiply equation (1) with

$p$ and equation (2) with

$q$ ,

$p^{2}x+pqy=p^{2}-pq$ ........... (3)

$q^{2}x-pqy=pq+q^{2}$ ............ (4)

Add (3) and (4),

$p^{2}x+q^{2}x=p^{2}+q^{2}$

$x\left ( p^{2}+q^{2} \right )=p^{2}+q^{2}$

$\Rightarrow x=1$

Put

$x=1$ in equation (1) to get the value of

$y$ ,

$p(1)+qy=p-q$

$p+qy=p-q$

$qy=-q$

$\Rightarrow y=-1$

Solve
$2x+3y=8$
$2x=2+3y$

$2x+3y = 8$

$2x+2 = 3y$

$2x+3y = 8$

$\Rightarrow 2x-3y = 2$

__________________

$4x = 10.\Rightarrow x = 5/2$

$2.5/2+3y = 8$

$\Rightarrow 3y = 3 \Rightarrow y = 1$

$\therefore x = 5/2, y = 1$

The sum of two numbers is $80$ and the greater number exceed twice the smaller by $11$ . Find the larger number.

Solution:-

Let the two numbers be

$x$ and

$y$ such that

$x > y$ ..

Now according to the question-

Condition I:-

$x + y = 80 ..... \left( 1 \right)$

Condition II:-

$x = 2y + 11 ..... \left( 2 \right)$

Substituting the value of

$x$ in

${eq}^{n} \left( 1 \right)$ we have

$\left( 2y + 11 \right) + y = 80$

$3y + 11 = 80$

$3y = 80 - 11$

$y = \cfrac{69}{3} = 23$

Substituting the value of

$y$ in

${eq}^{n} \left( 2 \right)$ , we have

$x = 2 \times 23 + 11$

$\Rightarrow x = 46 + 11 = 57$

Hence the required numbers are

$57$ and

$23$ .

Solve: $3a+5b=26\\ a+5b=22$

$3a+5b=26$ (1)

$a+5b=22$ (2)

Subtracting (2) from (1):

$2a = 4\implies a = 2$ (3)

Substituting (3) in (2):

$2+5b=22\implies 5b=20\implies b=4$

$\boxed{a=2, b=4}$

Solve by elimination method
$3x+4y=-25$ ;
$2x-3y=6$

Multiplying first equation by 2

$6x+8y=-50$

Multiply second equation by 3

$6x-9y=18$

Subtracting second equation from first equation,

$17y=-68$

$y=-4$

Substituting value of y in

$2x-3y=6$ ,

$x=-3$

If x=a , y=b is the solution of the pair of equations $x-y=2$ and $x+y=4$ , find the values of a and b.

Given pair of linear equations-

$x - y = 2 ..... \left( 1 \right)$

$x + y = 4 ..... \left( 2 \right)$

Adding equation

$\left( 1 \right)$ and

$\left( 2 \right)$ , we have

$\left( x - y \right) + \left( x + y \right) = 2 + 4$

$x - y + x + y = 6$

$2x = 6$

$\Rightarrow x = \cfrac{6}{2} = 3$

Substituting the value of

$x$ in equation

$\left( 1 \right)$ , we have

$3 - y = 2$

$\Rightarrow y = 3 - 2 = 1$

Hence the values of

$a$ and

$b$ are

$1$ and

$2$ respectively.

Solve the following pair of linear equations by the substitution method.
$0.2x+0.3y=1.3 \\ 0.4x+0.5y=2.3$

0.2x+0.3y=1.3 ......(1)

0.4x+0.5y=1.3 ......(2)

Derive the value of y from equation (1)

$y = \frac{{1.3 - 0.2x}}{{0.3}}$ ......(3)

Substitute value of y in equation(3)

$\begin{array}{c}0.4x + 0.5\left( {\frac{{1.3 - 0.2x}}{{0.3}}} \right) = 2.3\\0.12x + 0.5\left( {1.3 - 0.2x} \right) = 0.69\\0.12x + 0.65 - 0.1x = 0.69\\0.02x = 0.64\\x = 32\end{array}$

Substitute value of x in equation (3)

$\begin{array}{c}y = \frac{{1.3 - 0.2\left( {32} \right)}}{{0.3}}\\ = \frac{{1.3 - 6.4}}{{0.3}}\\ = - 17\end{array}$

So, x=32 and y=-17

The number of solutions for $2 x + 3 y = 10$ and $4x+6y=25$ are

$2x + 3y = 10 \, ...(1)$

$4x + 6y = 25 \,...(2)$

$(1) \times 2 \Rightarrow 4x + 6y = 20$

Subtracting the last two equations,

we get

$0=5$ which is not possible.

Hence, there are no solutions for this system of equations.

Solve:
$50x + 24y = 1000$
$x + y = 100$

$50x+24y=1000$ .......(1)

$x+y=100$ ........(2)

Eqn

$(1)-50\times(2)$

$\Rightarrow -26y=-4000$

$y=\dfrac{2000}{13}$

$x=100-\dfrac{2000}{13}=\dfrac{-700}{13}$

Solve :
$\dfrac{4}{x}+\dfrac{5}{y}=7;\ \dfrac{3}{x}+\dfrac{4}{y}=5$

$\dfrac{4}{x}+\dfrac{5}{y} = 7$

$\dfrac{3}{x}+\dfrac{4}{y} = 5$

Let

$u = \dfrac{1}{x}$ and

$v = \dfrac{1}{y}$

$\therefore 4u+5v = 7$ (i)

$3u+4v = 5$ (ii)

(i)

$\times$ 3 & (ii)

$\times 4\times -1$ and perform their addition

$\Rightarrow$

$12u+15v = 21$

$12u-16v = -20$

_________________

$-v = 1$

$v = -1$

putting

$v = -1$ in (ii)

$3u+4(-1) = 5$

$3u = 9$

$u = 3$

$\therefore$ substituting the values of u and v

$x = \dfrac13$ and

$y = -1$

Solve the following pair of equations:
$13x+11y=70, 11x+13y=74$

$13x+11y=70$

$............(1)$

$11x+13y=74$

$............(2)$

Multiply by

$13$ in equation (1) and multiply by

$11$ in equation (2), we get

$169x+143y=910$

$............(3)$

$121x+143y=814$

$............(4)$

On subtracting equation (4) from (3), we get

$x=2$

On putting the value of

$x$ in equation (1), we get

$13\times 2+11y=70$

$11y=44$

$y=4$

Hence, this is the answer.

$\dfrac{x}{10}+\dfrac{y}{5}-1=0$ $\dfrac{x}{8}+\dfrac{y}{6}=15$
Solve By method elimination

$\cfrac { x }{ 10 } +\cfrac { y }{ 5 } =1$

$\cfrac { x }{ 60 } +\cfrac { y }{ 30 } =\cfrac { 1 }{ 6 }$ (Multipying by

$\cfrac{1}{6}$ on both hand.)

$\longrightarrow (1)$

$\cfrac { x }{ 8 } +\cfrac { y }{ 6 } =15$

$\cfrac { x }{ 40 } +\cfrac { y }{ 30 } =\cfrac { 15 }{ 5 }=3$ (Multipying by

$\cfrac{1}{5}$ on both hand.)

$\longrightarrow (2)$

Subtracting

$(1)$ from

$(2)$ ,

$\cfrac{x}{40}-\cfrac{x}{60}=3-\cfrac{1}{6}=\cfrac{17}{6}$

$\cfrac{20x}{40\times 60}=\cfrac{17}{6}$

$x=340$

So,

$\cfrac{340}{10}+\cfrac{y}{5}=1\Longrightarrow y=-165$

Solve the following simultaneous equation by elimination method :
$3x-4y=10;\quad 4x+3y=5$ .

1198602_1229288_ans_022b256ca7b041fb977420d8fd6ccead.jpg

Solve for $x,y$
$4x-y=4;x-4y=16$

$4x - y = 4, \, ....(1) \,\, x - 4y = 16 \, ...(2)$

$(1) - 4 (2)$

$\Rightarrow \,\, 4x - y \,\,\,\,= 4$

$4x - 16y = 64$

$-$

________________

$15 y = -60$

$y = -4$

$x = 0$

Solve using the method of substitution
2x - 3y = 7 ; 5x + y = 9

$2 x - 3 y = 7$ ----------------

$(1)$

$5 x + y = 9$

$y = 9 - 5 x$ -------------

$(2)$

Now

$,$ substitute equation

$(2)$ in equation

$(1)$

Then

$, 2 x - 3(9 - 5 x) = 7$

$2 x - 27 + 15 x = 7$

$17 x = 27 + 7$

$17 x = 34$

$x = 34 / 17 = 2$

$y = 9 - 5 \times 2 = 9-10 = -1$

Therefore

$x$ is

$2$ and y is

$-1$

Solve for $x$ and $y$ : $\dfrac{x}{2}+\dfrac{y}{4}=3, 2x-y=4$ .

$\dfrac{x}{2}+\dfrac{y}{4}=3)\times 4\Rightarrow 2x+y=12$

$2x-y=4$

_______________

$4x=16\Rightarrow x=4$

$y=2x-4=8-4$

$\Rightarrow y=4$ .

1214858_1396571_ans_96bf0f0c9c4a4f3ca5060c501c15b2e6.jpg

Solve the following simultaneous equation:
(a) $x+y=4,x-y=6$

Given,

$x+y=4$ ....(1) and

$x-y=6$ .....(2).

Now adding (1) and (2) we get,

$2x=10$ or,

$x=5$ .

Again (1)

$-$ (2) gives,

$2y=-2$

or,

$y=-1$ .

Solve by elimination method: $13x + 11y = 70, 11x + 13y = 74$ .

$13x+11y=70---eqn(1)$

$11x+13y=74---eqn(2)$

Multiplying

$eqn(1)$ with

$11$

$11(13x+11y)=11\times70$

$\implies 143x+121y=770 ---eqn(3)$

Multiplying

$eqn(2)$ with

$13$

$13(11x+13y)=13\times74$

$143x+169y=962---eqn(4)$

Subtracting

$eqn(3)$ from

$eqn(4)$ we get,

$48y=192$

$\implies y=4$

Now let us eliminate y.

Multiplying

$eqn(1)$ with 13 then we get,

$169x+143y=910---eqn(5)$

Multiplying

$eqn(2)$ with 11, we get

$121x+143y=814----eqn(6)$

Subtracting

$eqn(6)$ from

$eqn(5)$ ,we get

$x=2$

$x=2$ and

$y=4$ is our solution

Solve $8x + 3y = 14 ; 3x + y = 5$ by using substitution method.

We have

$8x+3y=14$ ..........(i)

$3x+y=5$ .............(ii)

from (ii) we can write

$y=5-3x$

putting

$y=5-3x$ in (i) we get

$8x+3(5-3x)=14$

$8x+15-9x=14$

$-x=-1$

$x=1$

put

$x=1$ in

$y=5-3x$

$y=5-3=2$

Hence

$x=1$ and

$y=2$

Solve the equations :
$2x-y=2$
$4x-y=4$

$2x-y=2$ ........(1)

$4x-y=4$ ........(2)

Subtracting equation (1) from equation (2), we get,

$2x=2$

$x=1$

Put

$x=1$ in

$(1)$ w get

Therefore,

$y=0$

Draw the graph of the equation $x=0,\,y=0$

$x=0$ is the graph of

$y-$ axis

$\therefore$ the graph of

$x=0$ is the line

$YO{Y}^{\prime}$

$y=0$ is the graph of

$x-$ axis

$\therefore$ the graph of

$y=x=0$ is the line

$XO{X}^{\prime}$

1342374_1363417_ans_095ae7a2ccfc437ba255bdd496d34f53.PNG

Solve
$x-2y=-1,2x-y=7$

$x-2y=-1$ …………..

$(1)$

$2x-y=7$ ………….

$(2)$

Multiplying equation

$(1)$ by

$2$ and subtract equation

$(2)$ from it

$2x-4y=-2$

$2x-y=7$

_____________

$-3y=-9$

____________

$y=3$

Substitute the value of

$y$ in equation

$(1)$

Then

$x=2y-1=6-1=5$

$[x=5]$ .

Solve :
$3x-2y=1$
$2x+y=3$

$3x-2y=1\\2x+y=3\\y=3-2x\\3x-2(3-2x)=1\\3x-6+4x=1\\7x=7\\x=1\\3(1)-2y=1\\1-2y=1\\2y=2\\y=1$

Solve $x+y=14$ , $x-y=4$

$x-y=14\\x+y=4$

Add the two equations,

$\Rightarrow 2x=18\\x=9,\\y=4-x\\=4-9\\=-5$

$\therefore x=9,y=-5$

Find the solution of $x+2y=10$ and $2x+4y=8$ graphically.

Given,

$x + 2y = 10 \quad$

$\Rightarrow x=10-2y$

For

$y=3$

$x=10-2(3)=4$

For

$y=4$

$x=10-2(4)=2$

$x$	$4$	$2$
$y$	$3$	$4$

For

$2x + 4y = 8 \quad$

$\Rightarrow x + 2y = 4$

$\Rightarrow x=4-2y$

For

$y = 1$

$x=4-2(2)=2$

For

$y = 3$

$x = 4-2(3)=-2$

$x$	$2$	$-2$
$y$	$1$	$3$

The graph

$x + 2y = 10$ cuts the

$x$ and

$y$ -axis at

$\left( 10, 0 \right)$ and

$\left( 0, 5 \right)$ respectively, while the graph of

$2x + 4y = 8$ cuts the

$x$ and

$y$ -axis at

$\left( 4, 0 \right)$ and

$\left( 0, 2 \right)$ respectively.

Since the graph consists of a pair of parallel lines, the system of equations has no solution.

1207202_1536659_ans_30ad2bed49024d15bf2a889408dce79c.png

Solve : $x+y=110$ , $y=x+4$ .

$x+y=110, y=x+4$

substituting

$y=x+4$ in

$x+y=110$

$\Rightarrow x+(x+4)=100$

$\Rightarrow 2x=110-4=106$

$\therefore x=\frac{106}{2}=53$

Put

$x=53$ in

$y=x+4$

$\Rightarrow y=53+4=57$

$\therefore x=53, y=57$

Check whether the following equations are consistent or inconsistent , solve them graphically.
x - y = 8 , 3x - 3y = 16

x - y = 8 , 3x - 3y = 16

$\dfrac{a_{1}}{a_{2}} = \dfrac{1}{3} ,\dfrac{b_{1}}{b_{2}}=\dfrac{-1}{-3}= \dfrac{1}{3} \dfrac{c_{1}}{c_{2}}=\dfrac{8}{16}=\dfrac{1}{2}$

$\therefore \dfrac{a_{1}}{a_{2}} = \dfrac{b_{1}}{b_{2}} \dfrac{c_{1}}{c_{2}}$

$\therefore$ the linear equation are consistent

$\therefore$ They have infinity solutions.

x y = 8 x +3y = 16

x y (x ,y) x y (x,y)0

-8 (0,-8) 2

$- \dfrac{10}{3}$

1796650_1836181_ans_5e470565c80248539ae4e359873b94ce.png

Solve, graphically, the following pairs of equations:
$x - 5 = 0$
$y + 4 = 0$ .

$x - 5 = 0\Rightarrow x = 5$

$y + 5 = 0\Rightarrow y = -4$
Following is the graph of the two equations

$x = 5$ and

$y = -4$ .
1800721_1843781_ans_2901f8feb05641b7910cedd05c80e3d8.png

Solve, graphically, the following pairs of equations:
$2x + y = 23$
$4x - y = 19$ .

$2x + y = 23\Rightarrow y = 23 - 2x$
The table for

$y = 23 - 2x$ is

$x$	$5$	$10$	$15$
$y$	$13$	$3$	$-7$

Also, we have

$4x - y = 19$

$\Rightarrow y = 4x - 19$
The table for

$y = 4x - 19$ is

$x$	$3$	$4$	$6$
$y$	$-7$	$-3$	$5$

Intersecting point

$= (7, 9)$ .
1800725_1843782_ans_e9a40991a23c4598b501635507e0a2cd.png

Use the graphical method to find the value of $k$ , if $(k, -3)$ lies on the straight line $2x + 3y = 1$ .

$2x + 3y = 1$

$\Rightarrow 3y = 1 - 2x$

$\Rightarrow y = \dfrac {1 - 2x}{3}$
The table for

$2x + 3y = 1$ is

$x$	$-1$	$2$	$5$
$y$	$1$	$-1$	$-3$

Hence, from the graph, we can conclude that

$k = 5$ .

1800714_1843777_ans_4ba4cd0a7f45436098ace0978998c35f.png

Some people collected money to be donated in some orphanages. A part of the donation is fixed and the remaining depends on the number of children in the orphanage. If they donated Rs. $9500$ , in the orphanage having $50$ children and Rs. $13,250$ with $75$ children, find the fixed part of the donation and the amount donated for each child.
What values do these people possess?

Let the fixed donation be Rs.

$x$ and amount donated for each child be Rs.

$y$ .
Then, According to the question,

$x+50y=9500$ ....(i)

$x+75y=13250$ ....(ii)

Subtrating (i) from (ii) we get

$25y=3750$ or

$y=150$
From (i)

$x=9500-50y=9500-50\times 150=2000$
fixed amount donated

$=Rs.2000$
Amount donated for each child

$=Rs.150$

They possess values like Helpfulness, cooperation, happiness, caring.

Use the graphical method to find the value of $k$ , if $(5, k - 2)$ lies on the straight line $x - 2y + 1 = 0$ .

$x - 2y + 1 = 0$

$\Rightarrow 2y = x + 1$

$\Rightarrow y = \dfrac {x + 1}{2}$
The table for

$x - 2y + 1 = 0$ is

$x$	$1$	$3$	$5$
$y$	$1$	$2$	$3$

Hence, from the graph, we can conclude that,

$k - 2 = 3$

$k = 5$ .

1800720_1843779_ans_f860cb1c909e4b1e8cee2a9d071e404c.png

The force exerted to pull a cart is directly proportional to the acceleration produced in the body. Express the statement as linear equation in two variables and draw the graph of the same by taking the constant mass equal to 6 kg. Read from the graph, the force required when the acceleration produced is (i) $5$ $m/sec^{2}$ , (ii) $6$ $m/sec^{2}$

We have

$y$

$x$

Where

$y$ denotes the force,

$x$ denotes the acceleration and

$m$ denotes the constant mass.

Taking

$m = 6kg,$ we get

$y = 6x$

Now, we form a table as following the value of

$y$ below the corresponding value of

$x.$

$\frac{x}{y}\frac{0}{0}\frac{1}{6}\frac{2}{12}$

Plot the points

$(0,0), (1,6)$ and

$(2,12)$ on a graph paper and join any two points and obtain a line.

From the graph we see that if the acceleration is

$5 m/sec^2$ the force is 30 N and if the accleration is

$6 m/see^2$ , then the force is 36 N.

Using the same axes of co-ordinates and the same unit, solve graphically:
$x + y = 0$ and $3x - 2y = 10$ .
(Take at least $3$ points for each line drawn).

$x + y = 0$

$y = -x$
The table of

$x + y = 0$ is

$x$	$5$	$2$	$-5$
$y$	$-5$	$-2$	$5$

$3x - 2y = 10\Rightarrow x = \dfrac {10 + 2y}{3}$
The table of

$3x - 2y = 10$ is

$x$	$4$	$6$	$2$
$y$	$1$	$4$	$-2$

Intersection point

$= (2, -2)$
i.e.,

$x = 2$ and

$y = -2$ .
1800814_1843852_ans_37bd682084044fd086f6998485547756.png

Use the graphical method to find the value of $'x'$ for which the expressions $\dfrac {x + 2}{2}$ and $\dfrac {3}{4}x - 2$ are equal.

$y = \dfrac {3x + 2}{2}$
The table for

$y = \dfrac {3x + 2}{2}$ is

$x$	$2$	$4$	$-2$
$y$	$4$	$7$	$-2$

$y = \dfrac {3}{4} x - 2$
The table for

$y = \dfrac {3}{4}x - 2$ is

$x$	$4$	$-4$	$8$
$y$	$1$	$-5$	$4$

$x = -4$ .
1800828_1843854_ans_b20abf87a73d4ed5a36da3a2471632f8.png

Solve, graphically, the following pairs of equations:
$\dfrac {x + 1}{4} = \dfrac {2}{3}(1 - 2y)$
$\dfrac {2 + 5y}{3} = \dfrac {x}{7} - 2$ .

$\dfrac {x + 1}{4} = \dfrac {2}{3} (1 - 2y)$

$\Rightarrow \dfrac {x + 1}{4} = \dfrac {2}{3} - \dfrac {4y}{3}$

$\Rightarrow 12\times \dfrac {x + 1}{4} = 12\times \dfrac {2}{3} - 12\times \dfrac {4y}{3}$

$\Rightarrow 3(x + 1) = 8 - 16y$

$\Rightarrow 3x + 3 = 8 - 16y$

$\Rightarrow 3x + 3 - 8 = -16y$

$\Rightarrow 3x - 5 = -16y$

$\Rightarrow x = \dfrac {5 - 16y}{3}$
The table for

$\dfrac {x + 1}{4} = \dfrac {2}{3} (1 - 2y)$ is

$x$	$7$	$-9$	$23$
$y$	$-1$	$2$	$-4$

Also, we have

$\dfrac {2 + 5y}{3} = \dfrac {x}{7} - 2$

$\Rightarrow 21 \times \dfrac {2 + 5y}{3} = 21\times \dfrac {x}{7} - 21\times 2$

$\Rightarrow 7(2 + 5y) = 3x - 42$

$\Rightarrow 14 + 35y = 3x - 42$

$\Rightarrow 3x = 14 + 35y + 42$

$\Rightarrow 3x = 56 + 35y$

$\Rightarrow x = \dfrac {56 + 35y}{3}$
The table for

$\dfrac {2 + 5y}{3} = \dfrac {x}{7} - 2$ is

$x$	$7$	$-28$	$42$
$y$	$-1$	$-4$	$2$

Intersection point =

$(7, -1)$ .
1800730_1843787_ans_17cd1cf3032a43279cef04150b0f1ec5.png

Draw the graph of the pair of equations 2x + y = 4 and 2x -y =Write the vertices of the triangle formed by these lines and the y-axis.

Clearly from the graph the vertices of formed triangle are

$(0,0), (0,4), (2,0)$

By the graphical method, find whether the following pair of equations are consistent or not. If consistent, solve them.
$x + y = 3$ and $3x + 3y = 9$

Table of eq1

$x$	1	2	3
$y$	2	1	0

Table of eq2

$x$	1	2	3
$y$	2	1	0

The given equation can be written as

$x+y-3=0$ .......eq1 and

$3x+3y-9=0$ .........eq2
Here,

$\cfrac{a_1}{a_2}=\cfrac{1}{3},\cfrac{b_1}{b_2}=\cfrac{1}{3},\cfrac{c_1}{c_2}=\cfrac{3}{9}=\cfrac{1}{3}$

$\therefore \cfrac{a_1}{a_2}=\cfrac{b_1}{b_2}=\cfrac{c_1}{c_2}$
Thus, the given pair of equation is consistent.

Solve the following pair of equations:
$x+y=3.3 \\ \cfrac{0.6}{3x-2y}=-1;\ 3x-2y \neq 0$

Given equations are

$x+y=3.3$

$\cfrac{0.6}{3x-2y}=-1, 3x-2y \neq 0$

Rewriting the given equations, we get

$x+y=3.3$

multiplying both sides by

$10$

$\Rightarrow 10x+10y=33...................(1)$

and

$\cfrac{6}{10(3x-2y)}=-1$

$\Rightarrow 30x-20y=-6..................(2)$

On multiplying

$(1)$ by

$3$ and

$(2)$ by

$1$ , we get

$30x+30y=99$

$\underline {\underset {-}30x\underset {+}{-}20y=\underset{+}{-}6}$

$50y=105$

$\therefore y=2.1$

On putting

$y=2.1$ in

$(1)$ , we get

$10x+10\times2.1=33$

$\Rightarrow 10x+21=33$

$\Rightarrow 10x=33-21$

$\Rightarrow 10x=12$

$\Rightarrow x=1.2$

Hence, the solution is

$x=1.2,y=2.1$

Draw the graphs of $x + 3y + 3 = 0$ and $3x - 2y + 6 = 0.$ Plot only three points per line and mark the intersection point.

$x+3y+3=0$

$\implies x=-3-3y$

Let,

$y=0,$ then

$x=-3-3(0)=-3$

Similarly, when

$y=-1, x=0$ and when

$y=1, x=-6$

$\therefore$ the coordinates are

$(-3,0), (0,-1), (-6,1)$

Now,

$3x-2y+6=0$

$\implies 2y=3x+6$

$\implies y=\dfrac{3x+6}{2}$

Let,

$x=0$ ,

$y=3$

Similarly, when

$x=-2, y=0,$ and

$x=2, y=6$

$\therefore$ the coordinates are

$(0,3),(-2,0),(2,6)$

$\therefore$ the intersecting point is marked with the dark black dot

A purse contains only 25 paise and 10 paise coins. The total amount in the purse is Rs. 8.If the number of 25 paise coins is one-third the number of 10 paise coins, then the total number of coins in the purse is:

Let

$x$ be the number of

$25$ -paise coins and

$y$ be the number of

$10$ -paise coins.

Value of one

$25$ paise coin

$=Rs.\ 0.25$

Value of one

$10$ paise coin

$=Rs.\ 0.10$

Given that the total amount in the purse is

$Rs. \ 8.25$ .

$\therefore\ 0.25 x + 0.10 y = 8.25$

$\Rightarrow \cfrac 14x + \cfrac {1}{100}y=8.25$ ....(1)

Also given that, the number of

$25$ paise coins are one-third of

$10$ paise coins.

$\therefore \ x = \cfrac{1}{3}y$ ....(2)

Substituting (2) in (1), we get:

$\cfrac {1}{12}y+\cfrac {1}{100}y =\cfrac {825}{100}$

$\Rightarrow \cfrac {112}{1200}y = \cfrac {825}{100}$

$\Rightarrow y = \cfrac {825 \times 12}{112}$

$\Rightarrow y = 45$

Substituting value of

$y$ in (2), we get:

$\therefore x=15$

So, the total number of coins

$= 15 + 45 = 60$ .

Hence, there are a total of

$60$ coins in the purse.

A train covers a certain distance at a uniform speed. If the train could have been $10 km/hr$ faster, it would have taken $2$ hours less than the scheduled time. And if the train were slower by $10 km/hr$ , it would have taken $3$ hours more than the scheduled time. Find the distance covered by the train.

Let the distance covered by the train

$= d$ km and speed

$= s$ km/hr.

Distance

$=$ Speed

$\times$ Time

$=$ constant

where, Time is in hrs

$st = (s + 10) (t - 2)$ ......(i)

$st = (s - 10) (t + 3)$ .....(ii)
Simplifying the (i),

$st = st - 2s +10t - 20$

$\Rightarrow 2s - 10t = - 20$ .....(iii)
Simplifying the (ii) we get,

$st = st + 3s + 10t - 30$

$\Rightarrow 3s - 10t = 30$ ............(iv)
Multiplying (iii) by

$3$ & (iv) by

$2$ , we get

$6s - 30t = -60$ ..........(v)

$6s - 20t = 60$ .......... (vi)
Subtracting equation (vi) from (v), we get

$- 10t = -120 t = 12 hrs$
Putting the value of

$t$ in equation (iii),

$s = 50 km/hr$

$d = st = 50 \times 12 = 600$ km

Find the number of solutions for $2x-3y=5$ and $4x-6y=7$ .

$2x-3y-5=0\Rightarrow a_1=2,\,b_1=-3, c_1=-5$

$4x-6y-7=0\Rightarrow a_2=4,\,b_2=-6,\,c_2=-7$

$\Rightarrow \dfrac{a_1}{a_2}=\dfrac{2}{4}=\dfrac12;\,\dfrac{b_1}{b_2}=\dfrac{-3}{-6}=\dfrac12;\,\dfrac{c_1}{c_2}=\dfrac{-5}{-7}=\dfrac57$

$\therefore \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\ne\dfrac{c_1}{c_2}$
Therefore, the given system of equations are inconsistent and have no solution.
No. of solutions

$=0$

Solve the following system of equations
$\displaystyle \frac{6}{x+y}=\frac{7}{x-y}+3,\frac{1}{2\left ( x+y \right )}=\frac{1}{3\left ( x-y \right )}$ where $\displaystyle x+y\neq 0\: \: and\: \: x-y\neq 0$

$\displaystyle \frac{6}{x+y}=\frac{7}{x-y}+3$ ......(i)

Put x + y = a, x - y = b

Now the given equation reduces to

$\displaystyle \frac{6}{a}=\frac{7}{b}+3$ ........(ii)

Another equation is

$\displaystyle \frac{1}{2\left ( x+y \right )}=\frac{1}{3\left ( x-y \right )}$ .......(iii)

$\displaystyle \frac{1}{2a}=\frac{1}{3b}$

or a =

$\displaystyle \frac{3b}{2}$ ........(iv)

Put (iv) in (ii)

$\displaystyle \frac{6}{3b}\times 2=\frac{7+3b}{b}$

$4 = 7 + 3b$

$b = -1$ ......(v)

Put (v) in (iv)

$\displaystyle a=-\frac{3}{2}$

From our four assumptions

a = x + y =

$\displaystyle -\frac{3}{2}$

$b = x - y = -1$

Adding the two equations given above

$\displaystyle 2x=-\frac{5}{2}\Rightarrow x=-\frac{5}{4}$

$\displaystyle \therefore y=-\frac{5}{4}+1\Rightarrow y=-\frac{1}{4}$

The owner of a milk store finds that, he can sell $980$ litres of milk each week at Rs $14$ /litre and $1220$ litres of milk each week at Rs $16$ /litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs $17$ /litre

Assuming selling price per litre along the x-axis and demand along the

$y$ -axis. we have two points

$(14,980)$ and

$(16,1220)$ in the

$XY$ plane that satisfies the linear relationship between the selling price and demand.

Therefore, the linear relationship between selling price per litre and demand is the equation of the line passing through points

$(14,980)$ and

$(16,1220)$ , which is given by,

$\displaystyle y-980= \frac{1220-980}{16-14}(x-14)$

$\displaystyle y-980= \frac{240}{2}(x-14)$

$\displaystyle y-980= 120(x-14)$

i.e

$\displaystyle y= 120(x-4)+980$

when

$x\displaystyle =$ Rs

$17$ /litre,

$y\displaystyle =120(17-14)+980$

$\displaystyle \Rightarrow y= 120\cdot 3+980= 360+980= 1340$

Thus the owner of the milk store could sell

$1340$ litres milk weekly at Rs

$17$ /litre

A trust fund has $Rs. 30,000$ that must be invested in two different types of bonds. The first bond pays $5$ % interest per year, and the second bond pays $7$ % interest per year. Determine how to divide $Rs. 30,000$ among the two types of bonds. If the trust fund must obtain an annual total interest of
(a) $Rs. 1800$
(b) $Rs. 2000$

Given that first bond pays

$5$ % interest per year and second bond pays

$7$ % interest per year let the money invested in first bond be

$x$ and in second bond be

$y=30000-x$

therefore total interest

$=(5/100 \times x)+(7/100 \times (30000-x))=2100-0.02 \times x$

a.)given total interest

$=1800$

i.e.

$2100-0.02 x=1800$

therefore

$x=15000,y=30000-x=15000$

b.)given total interest

$=2000$

i.e.

$2100-0.02 x=2000$

therefore

$x=5000, y=30000-x=25000$

Father says to son "I am 9 times as old as you were when i was as old as you are". If the sum of their present ages is 50 years. What is the age of father ?

Let the present age of father and son be x and y respectively

Then as per question

$x+y=50$ ........................(1)

Given

$x=9y$ .......................................................(2)

substituting equation (2) in equation (1) we get

$9y+y=50$

$10y=50$

$y=5$ .........................................................................(3)

Putting value of y in eqn 2 we get

$x=9(5)=45$

We get

$x=45$

Then age of father is

$45$ years

In countries like USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius:
$F=\left(\dfrac{9}{5}\right)C+32$
(i) Draw the graph of the linear equation above using Celsius for $x$ -axis and Fahrenheit for $y$ -axis.
(ii) If the temperature is $30^oC$ , what is the temperature in Fahrenheit?
(iii) If the temperature is $95^oF$ , what is the temperature in Celsius?
(iv) If the temperature is $0^oC$ , what is the temperature in Fahrenheit and if the temperature is $0^oF$ , what is the temperature in Celsius?
(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.

(i) Since the equation between

$F$ and

$C$ is given, the graph can be drawn as shown above.

(ii) Given

$c=30$ , so

$F=\dfrac { 9 }{ 5 } \times 30+32=86 \ F$ .

(iii) Given

$F=95$ , so

$C=\dfrac { 5 }{ 9 } \times (F-32)=35 \ C$ .

(iv) Given

$C=0$ , so

$F=32$ and if

$F=0$ , we get

$C=-\dfrac{160}9$ .

(v) Put

$F=C=x$ in given equatin, we get

$\dfrac { 4x }{ 5 } =-32$ , from which we get

$x=-40$

Yes, there is a temperature which is numerically the same in both fahrenheit and celsius, the numerical value is

$-40$ .

The cost of $2\ kg$ of apples and $1\ kg$ of grapes on a day was found to be $Rs \ 160$ . After a month, the cost of $4\ kg$ of apples and $2\ kg$ of grapes is $Rs\ 300$ . Represent the situation algebraically and geometrically.

$x$	$65$	$55$	$45$	$35$
$y$	$20$	$40$	$60$	$80$

Let the cost of

$1$ kg of apples be

$x$ and that of

$1$ kg be

$y$

So the algebraic representation can be as follows:

$2x + y = 160$

$4x + 2y = 300 \Rightarrow 2x +y = 150$

The situation can be represented graphically by plotting these two equations.

$2x + y = 160 \Rightarrow y = 160-2x$

$x$	$70$	$60$	$50$	$40$
$y = 160-2x$	$20$	$40$	$60$	$80$

$2x + y = 150 \Rightarrow y = 150-2x$

$x$	$65$	$55$	$45$	$35$
$y = 150-2x$	$20$	$40$	$60$	$80$

We can see that the lines do not intersect anywhere, i.e. they are parallel. Hence we can not arrive at a solution.

On comparing the ratios $\dfrac{a_1}{a_2}, \dfrac{b_1}{b_2}$ and $\dfrac{c_1}{c_2}$ , find out whether the following pair of linear equations are consistent, or inconsistent.
(i) $3x + 2y = 5 ; 2x -3y = 7$

(ii) $2x -3y = 8 ; 4x -6y = 9$

(iii) $\dfrac{3}{2}x+\dfrac{5}{3}y=7 ; 9x -10y = 14$

(iv) $5x -3y = 11 ; -10x + 6y = -22$

v) $\dfrac{4}{3}x+2y=8 ; 2x + 3y = 12$

(i)

$\displaystyle \frac{a_1}{a_2} = \frac{3}{2} , \quad \frac{b_1}{b_2} = \frac{2}{-3}, \quad \frac{c_1}{c_2} = \frac{5}{7}$

$\because \displaystyle \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ ,

the lines intersect and have an unique consistent solution.

(ii)

$\displaystyle \frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}, \quad \frac{b_1}{b_2} = \frac{-3}{-6} = \frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{8}{9}$

$\because \displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ ,

the lines are parallel and have no solutions, i.e. the equations are an inconsistent pair

(iii)

$\displaystyle \frac{a_1}{a_2} = \frac{\frac{3}{2}}{9} = \frac{1}{6}, \quad \frac{b_1}{b_2} = \frac{\frac{5}{3}}{-10} = - \frac{1}{6}, \quad \frac{c_1}{c_2} = \frac{7}{14}= \frac{1}{2}$

$\because \displaystyle \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ ,

the lines intersect and have an unique consistent solution.

(iv)

$\displaystyle \frac{a_1}{a_2} = \frac{5}{-10} = -\frac{1}{2} , \quad \frac{b_1}{b_2} = \frac{-3}{6} = -\frac{1}{2}, \quad \frac{c_1}{c_2} = \frac{11}{-22} = - \frac{1}{2}$

$\because \displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ ,

the lines are coincident and have infinitely many solutions. The equations form a consistent pair of equations.

(v)

$\displaystyle \frac{a_1}{a_2} = \frac{\frac{4}{3}}{2} = \frac{2}{3}, \quad \frac{b_1}{b_2} = \frac{2}{3}, \quad \frac{c_1}{c_2} = \frac{8}{12} = \frac{2}{3}$

$\because \displaystyle \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ ,

the lines are coincident and have infinitely many solutions. The equations form a consistent pair of equations.

Draw the graphs of the equations $x- y + 1 = 0$ and $3x + 2y -12 = 0$ . Determine the coordinates of the vertices of the triangle formed by these lines and the $x$ -axis, and shade the triangular region.

$x-y+1=0$

$x-y=-1$ ...(i)

$3x+2y-12=0$

$3x+2y=12$ ...(ii)

$x-y=-1$

$y=1+x$

$x$	$0$	$-1$
$y$	$1$	$0$

Plot

$(0, 1)$ and

$(1, 2)$ on graph and join them to get eqn (i)

For

$3x+2y=12$

$\Rightarrow 2y=12-3x$

$\Rightarrow r=\dfrac{12-3x}{2}$

$x$	$0$	$4$
$y$	$6$	$0$

Plot point

$(0, 6)$ and

$(4, 0)$ on graph and join them to get eqn (ii)

$\triangle ABC$ is the required triangle with coordinates of

$A$ as

$(2, 3), B(-1, 0)$ and

$C(4, 0)$ .

Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) $10$ students of Class X took part in a Mathematics quiz. If the number of girls is $4$ more than the number of boys, find the number of boys and girls who took part in the quiz.
(ii) $5$ pencils and $7$ pens together cost Rs $50,$ whereas $7$ pencils and $5$ pens together cost Rs $46.$ Find the cost of one pencil and that of one pen.

i) Consider number of boys

$= x$

and number of girls

$= y.$

According to the question,

$x + y = 10$
or,

$y = 10 - x$ ………(i)

$y = x + 4$ ………..(ii)

$x$	$1$	$2$	$3$	$4$
$y$	$9$	$8$	$7$	$6$

$x$	$1$	$2$	$3$	$4$
$y$	$5$	$6$	$7$	$8$

Graphical solution : Girls

$= 7,$ boys

$= 3$

(ii) Consider price of a pencil is

$x$

and price of a pen is

$y.$

According to the question,

$5x + 7y = 50$ .....(i)

$7x + 5y = 46$ .….(ii)

$x$	$1$	$2$	$3$	$4$
$y$	$7.8$	$6.4$	$5$	$4.2$

$x$	$1$	$2$	$3$	$4$
$y$	$6.4$	$5.7$	$5$	$4.2$

Graphical solution :

Price of a pencil is Rs

$3$

Price of a pen is Rs

$5.$

Find the value of $k$ , if $x=2, y=1$ is a solution of the equation $2x+3y=k$ .

Equation

$2x+3y=k$

Where

$x=2,y=1$

Put the value of

$x$ and

$y$ in equation

$\Longrightarrow k=2\times 2+3\times 1$

$\Longrightarrow k=4+3$

$\therefore k=7$

Which one of the following options is true, and why?
$y=3x+5$ has
(i) a unique solution,
(ii) only two solutions,
(iii) infinitely many solutions

In the given two variable, if we put the value of

$x$ , then we get the corresponding value of

$y$

if value of

$x=0$ then

$\Rightarrow y=3\times 0+5$

$\Rightarrow y=5$

if value of

$x=1$ then

$\Rightarrow y=3\times 1+5$

$\Rightarrow y=8$

if value of

$x=2$ then

$\Rightarrow y=3\times 2+5$

$\Rightarrow y=11$

if value of

$x=-1$ then

$\Rightarrow y=3\times -1+5$

$\Rightarrow y=2$

Hence, all these pairs

$(0,5),(1,8),(2,11),(-1,2)$ are the solution for the given pair of equation

Hence, the given equation had infinitely many solutions.

Solve the following pair of linear equations by the substitution method.

(i) $x + y = 14; x -y = 4$

(ii) $s -t = 3; \dfrac{s}{3}+\dfrac{t}{2}=6$

(vi) $\dfrac{3x}{2}-\dfrac{5y}{3}=-2 ; \dfrac{x}{3}+\dfrac{y}{2}=\dfrac{13}{6}$

(i)

$x + y = 14 \Rightarrow y = 14-x$

Substituting this value in the second equation, we get

$x - (14-x) = 4$

$2x = 18 \Rightarrow x = 9$

Substituting this value of x in the first equation, we get

$9 +y = 14 \Rightarrow y= 5$

(ii)

$s - t = 3 \Rightarrow s = t+3$

Substituting in 2nd equation

$\displaystyle \frac{t+3}{3} + \frac{t}{2} = 6$

$\displaystyle \frac{2t+6+3t}{6} = 6$

$\displaystyle 5t+6 = 36 \Rightarrow t = 6$

Substituting value of t in 1st equation

$s = t+3 = 9$

(iii)

$\because \quad \dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$

Hence all the points lying on the line

$y = 3x-3$ like

$x = 2, y=3;$ are a solution.

(iv)

$0.2x +0.3y = 1.3 \Rightarrow x = \dfrac{1.3-0.3y}{0.2}$

Substituting this value in the second equation

$0.4 \times \dfrac{1.3-0.3y}{0.2} +0.5 y = 2.3$

$2.6 - 0.6y +0.5y = 2.3$

$2.6-2.3 = 0.6y-0.5y$

$0.1 y = 0.3 \Rightarrow y = 3$

Substituting this value of

$y,$

$x = \dfrac{1.3-0.3y}{0.2} = \dfrac{1.3 - 0.3 \times3}{0.2} = \dfrac{0.4}{0.2}$

$x = 2$

(v)

$\sqrt{2} x + \sqrt {3}y = 0 \Rightarrow y = - \dfrac{\sqrt{2}}{\sqrt{3}} x$

Substituting this in 2nd equation

$\sqrt3 x- \sqrt8 \times \left(- \dfrac{\sqrt{2}}{\sqrt{3}} x\right) = 0$

$3x +4x = 0 \Rightarrow x = 0$

Substituting value of

$x,$

$y = -\dfrac{\sqrt2}{\sqrt3} \times 0 = 0$

(vi)

$\displaystyle \frac{3x}{2} - \frac{5y}{3} = -2$

$\dfrac{5y}{3} = \dfrac{3x}{2} + 2 \Rightarrow y = \dfrac{9x+12}{10}$

Substituting this in 2nd equation

$\dfrac{x}{3} + \dfrac{9x +12}{20} = \dfrac{13}{6}$

$\dfrac{47x +36}{60} = \dfrac{13}{6}$

$47x + 36 = 130$

$47x = 94 \Rightarrow x = 2$

Substituting this value of

$x$ in 1st equation

$y = \dfrac{9x+12}{10} = \dfrac{9 \times 2 +12}{10} = 3$

Aftab tells his daughter, "Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be." (Isnt this interesting?) Represent this situation algebraically and graphically.

Consider Aftab's age as

$x$ and his daughter's age as

$y.$

Then , seven years ago,

Aftab's age

$= x -7$

His daughter's age

$=y-7$

According to the question,

$x - 7 = 7 \left ( y-7 \right )$

$x -7 = 7y - 49$

$x-7y = -49 +7$

$x-7y = - 42$ …(i)

After three years,
Aftab's age

$= x +3$
His daughter's age

$= y+3$
According to the question,

$x + 3 = 3 \left(y + 3\right)$

$x + 3 = 3 y + 9$

$x -3 y = 9- 3$

$x -3 y =6$ …(ii)

Representing equation

$(i)$ and

$(ii)$ geometrically, we plot these equations by finding points on the lines representing these two equations

$x-7y = - 42 \Longrightarrow x = 7y - 42$

$x = 7y-42$	$42$	$35$	$49$
$y$	$12$	$11$	$13$

$x - 3y = 6 \Longrightarrow x = 3y + 6$

$x = 3y+6$	$42$	$36$	$48$
$y$	$12$	$10$	$14$

From the graph we can see that two lines will intersect at a point.

$x-7y = - 42$

$x -3 y =6$

On subtracting the two equations, we get

$4y = 48$

$y = 12$

Substituting value of

$y$ in (2),

$x - 36 = 6$

$\therefore x= 42$

Form the pair of linear equations in the following problems, and find their solutions(if they exist) by the elimination method :

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces toIt becomes $\dfrac{1}{2}$ if we only add 1 to the denominator. What is the fraction?

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

(iii) The sum of the digits of a two-digit number isAlso, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

(iv) Meena went to a bank to withdraw Rs. $2000$ . She asked the cashier to give her Rs. $50$ and Rs. $100$ notes only. Meena got 25 notes in all. Find how many notes of Rs. $50$ and Rs. $100$ she received.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs. $27$ for a book kept for seven days, while Susy paid Rs. $21$ for the book she kept for five days. Find the fixed charge and the charge for each extra day.

(i) Let the fraction be

$\dfrac{x}{y}$

Given,

$\dfrac{x+1}{y-1} = 1$

$x+1 = y-1$

$x-y = -2$ .....(1)

Also,

$\dfrac{x}{y+1} = \dfrac{1}{2}$

$2x = y+1$

$2x-y=1$ .....(2)

Substracting (1) from (2)

$x = 3$

Substituting value of x in (2),

$6-y=1$

$y=5$

Hence, the fraction is

$\dfrac{3}{5}$

(ii) Let the age of Nuri be x and that of Sonu be y.

Given,

$\left(x-5\right) = 3 \left(y-5\right)$

$3y-x = 10$ ..... (1)

And,

$\left(x+10\right) = 2 \left(y+10\right)$

$2y -x = -10$ ....(2)

Subtracting (2) from (1), we get

$y=20$

Substituting value of y in (1)

$60-x = 10$

$x = 50$

Hence, the ages of Nuri and Sonu are 50 and 20 respectively.

(iii) Let the two digit number be

$10x+y$

Given,

$x+y = 9$ .....(1)

And,

$9 \left(10x+y\right) = 2\left(10y+x\right)$

$90x+9y = 20y +2x$

$88x-11y = 0$ .....(2)

Multiplying (1) by 11, we get

$11x + 11y = 99$ ....(3)

Adding (1) and (3)

$99x = 99$

$x = 1$

Substituting this value of x in (1),

$1+y = 9$

$y=8$

Hence, the number is 18.

(iv) Let the number of Rs 100 notes be x and the number of Rs 50 notes be y.

Given,

$x + y = 25$ ...(1)

And,

$100x + 50y = 2000$ ...(2)

Multiplying (1) by 50, we get

$50x + 50 y = 1250$ ....(3)

Subtracting (3) from (2),

$50x = 750$

$x = 15$

Substituting value of x in (1),

$15+y = 25$

$y=10$

(v)

$\text{Let the fixed charge for first 3days be}$

$Rs\,x$

$\text{and after 3days the charge for each day be}$

$Rs\, y$ .

Given,

$x +4y = 27$ ....(1) [first 3days + 4days]

And,

$x + 2y = 21$ ...(2) [first 3days + 2days]

Subtracting (2) from (1), we get

$2y = 6$

$y = 3$

Substituting value of y in (1)

$x +4 \times 3 = 27$

$x+12=27$

$x=15$

Hence, the fixed charge is Rs 15 and the per day charge is Rs 3.

Solve the following pair of linear equations:
$(i) \;\quad px + qy = p- q ; \;qx -py = p + q$
$(ii) \quad ax + by = c ; \;bx + ay = 1 + c$

(i)

$px + qy = p-q\quad\quad\dots(i)$

$qx - py = p+q\quad\quad\dots(ii)$

Multiplying

$(i)$ by p and

$(ii)$ by

$q$ we get

$(iii)$ and

$(iv)$ respectively as

$p^2x + pqy = p^2 - pq\quad\quad\dots(iii)$

$q^2x - pqy = pq + q^2\quad\quad\dots(iv)$

Adding

$(iii)$ and

$(iv)$ we get,

$(p^2 + q^2)x = (p^2 + q^2)$

$\Rightarrow x = 1$

Substituting value of

$x$ in

$(i)$ , we get

$p + qy = p - q$

$qy = -q$

$y = -1$

Hence,

$x = 1, \: y= -1$

(ii)

$ax + by = c\quad\quad\dots(i)$

$bx + ay = 1 +c\quad\quad\dots(ii)$

Multiplying

$(i)$ by

$b$ and

$(ii)$ by a we get

$(iii)$ and

$(iv)$ respectively as

$abx + b^2y = bc\quad\quad\dots(iii)$

$abx + a^2y = a +ac\quad\quad\dots(iv)$

Subtracting

$(iv)$ from

$(iii)$ , we get

$(b^2 - a^2) y = (b-a)c - a$

$y = \dfrac{\left( b-a \right)c - a}{\left(b^2 - a^2\right)}$

Substituting value of

$y$ in

$(i)$ , we get

$ax + b \times \dfrac{\left(b-a \right)c - a}{\left(b^2 - a^2 \right)} = c$

$ax = c - \dfrac{\left(b-a\right)bc -ab}{\left(b^2 - a^2\right)}$

$ax = \dfrac{c \left(b^2 - a^2\right) - b^2c+abc + ab}{\left(b^2 -a^2\right)}$

$x = \dfrac{cb^2 - ca^2 - b^2c + abc +ab}{a\left(b^2-a^2 \right)}$

$x = \dfrac{ab+abc-ca^2}{a\left( b^2-a^2\right)}$

$x = \dfrac{b-bc-ca}{\left(b^2-a^2\right)}$

$x = \dfrac{b +c \left(b-a\right)}{\left(b^2 - a^2\right)}$

Draw the graphs of the equations $5x- y = 5$ and $3x -y = 3$ . Determine the co-ordinates of the vertices of the triangle formed by these lines and the y axis.

For the equation

$3x-y=3$

$y=3x-3$

$x=0,\ y=-3$
For

$x=1,\ y=0$
For

$x=2,\ y=3$

$x$	$0$	$1$	$2$
$y$	$-3$	$0$	$3$

For the equation

$5x-y=5$

$\Rightarrow y=5x-5$

$x=0,\ y=-5$
For

$x=1,\ y=0$
For

$x=2,\ y=5$

$x$	$0$	$1$	$2$
$y$	$-5$	$0$	$5$

Plotting the equations on the graph, we get

$\Delta ACE$ with

$A (1,0),\ C (0,-3),$ and

$E (0,-5)$

Solve the following pair of linear equations by the elimination method and the substitution method :

(i) $x + y = 5$ and $2x -3y = 4$
(ii) $3x + 4y = 10$ and $2x -2y = 2$
(iii) $3x -5y -4 = 0$ and $9x = 2y + 7$
(iv) $\dfrac{x}{2}+\dfrac{2y}{3}=-1$ and $x-\dfrac{y}{3}=3$

(i)

$x+y=5$ ...(i)

$2x-3y=4$ ...(ii) (elimination method)

Multiply eqn (i) by

$2$

$2x+2y=10$

$2x-3y=4$

$- \,\,\,\,\,\,\,+ \,\,\,\,\, \,\,\,-$

____________

$5y=6$

$\Rightarrow y=\dfrac{6}{5}$

Substitute

$y=\dfrac{6}{5}$ in eqn (i)

$x+y=5$

$x+\dfrac{6}{5}=5$

$\Rightarrow x=\dfrac{5\times 5}{5}-\dfrac{6}{5}$

$=\dfrac{19}{5}$

Substitution method

$x+y=5$ ....(i)

$2x-3y=4$ ....(ii)

$y=5-x$ ...(iii)

Putting eqn. (3) in (2)

$2x-3(5-x)=4$

$2x-15+3x=4$

$x=\dfrac{19}{5}$

$y=\dfrac{5\times 5}{5}-\dfrac{19}{5}$

$y=\dfrac{6}{5}$

(ii)
Elimination method

$3x+4y=10$ ...(i)

$2x-2y=2$ ...(ii)

Multiply eqn (2) by 2

$4x-4y=4$

$3x+4y=10$

____________

$7x =14$

$x=2$

$2\times 2-2y=2$

$\Rightarrow 4-2=2y$

$y=1$

Substitution method

$3x+4y=10$ ...(i)

$2x-2y=2 \Rightarrow x-y=1$

$x-1=y$ ...(ii)

Putting eqn (ii) in (i)

$3x+4(x-1)=10$

$3x+4x-4\,\,\,\,=10$

_________________

$7x =14$

$x=2$

$y=x-1$

$y=2-1$

$\Rightarrow y=1$

(iii)

$3x-5y=4$ ...(i)

$9x-2y=7$ ...(ii) (elimination method)

Multiply eqn (i) by

$3$
(ii)-(i)

$\times 3$

$9x-2y=7$

$9x-15y=12$

_____________

$13 y = -5$

$\Rightarrow y=\dfrac{-5}{13}$

$3x-5\times \dfrac{-5}{13}=4$

$\Rightarrow 3x=4-\dfrac{25}{13}$

$3x=\dfrac{52-25}{13}$

$x=\dfrac{27}{13\times 3}$

$\Rightarrow x=\dfrac{9}{13}$

Substitution method

$3x-5y=4$ ...(i)

$9x-2y=7$

$\Rightarrow 9x-7=2y$

$\dfrac{9x-7}{2}=y$

$3x-5y=4$

$\Rightarrow 3x-5\left(\dfrac{9x-7}{2}\right)=4$

$\Rightarrow 3x-\dfrac{45}{2}x+\dfrac{35}{2}=4$

$\Rightarrow \dfrac{6x}{2}-\dfrac{45}{2}x=4-\dfrac{35}{2}$

$\Rightarrow \dfrac{-39x}{2} = \dfrac{8-35}{2}$

$x=\dfrac{27}{39}$

$\Rightarrow x=\dfrac{9}{13}$

Putting

$x=\dfrac{9}{13}$ in

$3x-5y=4$

$\dfrac{3\times 9}{13}-5y=4$

$\dfrac{27}{13}-5y=4$

$\Rightarrow \dfrac{27}{13}-4=5y$

$\Rightarrow \dfrac{27-52}{13}=5y$

$\dfrac{-25}{13} = 5y$

$\Rightarrow y=\dfrac{-5}{13}$

(iv)Elimination method

$\dfrac{x}{2}+\dfrac{2y}{3}=-1$

$\dfrac{3x+4y}{6}=-1$

$\Rightarrow 3x+4y=-6$ ...(i)

$x-\dfrac{y}{3} =3$

$\Rightarrow 3x-y=9$ ...(ii)

(i)

$-$ (ii)

$3x+4y=-6$

$3x-y=-9$

$-\,\,\,\, + \,\,\,\,\,\,\, -$

______________

$5y=-15$

$\Rightarrow y=-3$

Putting

$y=-3$ in eqn (i)

$3x-12=-6$

$\Rightarrow 3x=-6+12$

$\Rightarrow 3x=6$

$\Rightarrow x=2$

Substitution method

$3x+4y=-6$ ...(i)

$3x-y=9$ ...(ii)

$3x-9=y$ ....(iii)

Putting enq (iii) in (i)

$3x+4(3x-9)=-6$

$\Rightarrow 3x+12x-36=-6$

$\Rightarrow 15x=-6+36$

$\Rightarrow 15x=30$

$\Rightarrow x=2$

Putting x=2 in eq (iii)

$6-9=y$

$\Rightarrow y=-3$

Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method :

(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for $20$ days she has to pay $\text{Rs. } 1000$ as hostel charges whereas a student B, who takes food for $26$ days, pays $\text{Rs. } 1180$ as hostel charges. Find the fixed charges and the cost of food per day.

(ii) A fraction becomes $\dfrac{1}{3}$ when $1$ is subtracted from the numerator and it becomes $\dfrac14$ when $8$ is added to its denominator. Find the fraction.

i) Let the fixed monthly hostel charges be

$\text{Rs. }x$ and the cost of food per day taken from the mess be

$y$ .

As per the given condition,

$x+20y=1000$ ...(1)

$x+26y=1180$ ...(2)

Subtracting eq(1) from eq (2), we get

$\Rightarrow 6y = 180$

$\Rightarrow y = 30$

Substitute

$y=30$ in eq (1), we get

$\Rightarrow x+20(30) = 1000$

$\Rightarrow x+600 = 1000$

$\Rightarrow x=400$

So, the fixed monthly hostel charge is

$\text{Rs }600$ and the cost of food per day is

$\text{Rs }30$ .

ii) Let the numerator be

$x$ and the denominator be

$y$ , so the fraction will be

$\dfrac { x }{ y }$ .

$\dfrac{ x-1}{ y}=\dfrac { 1 }{ 3 }$

$\Rightarrow 3x-y=3\Rightarrow 3x-3=y$ ...(1)

$\dfrac { x }{ y+8}=\dfrac14$

$\Rightarrow 4x=y+8$ ...(2)

Substituting for

$y$ in (2) we get,

$4x=3x-3+8$

$x=5$

Putting

$x=5$ in eq(1), we get

$y=12$

$\therefore$ fraction

$=\dfrac{x}{y}=\dfrac{5}{12}$

Solve $2x + 3y = 11$ and $2x -4y = -24$ and hence find the value of m for which $y = mx + 3$ .

$2x + 3y = 11$

$2x = 11 - 3y$

Substituting this in 2nd equation, we get

$(11-3y) - 4y = -24$

$11-7y = -24$

$7y = 11+24$

$y = \dfrac{35}{7} = 5$

Substituting value of y

$2x = 11- 3 \times 5$

$x = \dfrac{-4}{2} = -2$

Now, to find the value of m, we substitute the values of x and y in the equation

$5 = m \times(-2) +3$

$m = \dfrac{5-3}{-2} = -1$

The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Let the number of rows be x and number of students in a row be y.

Total students of the class= Number of rows

$\times$ Number of students in a row

$= x\times y=xy$

Case 1

Total number of students

$= (x − 1) (y + 3)$

$\Rightarrow xy = (x − 1) (y + 3) = xy − y + 3x − 3$

$\Rightarrow 3x − y − 3 = 0$

$\Rightarrow 3x − y = 3$ ..... (i)

Case 2

Total number of students

$= (x + 2) (y − 3)$

$\Rightarrow xy = xy + 2y − 3x − 6$

$\Rightarrow 3x − 2y = −6$ ..... (ii)

Subtracting equation (ii) from (i),

$\Rightarrow (3x − y) − (3x − 2y) = 3 − (−6)$

$\Rightarrow − y + 2y = 3 + 6$

$\Rightarrow y = 9$

By substituting value of y in (i), we get

$\Rightarrow 3x − 9 = 3$

$\Rightarrow 3x = 9 + 3 = 12$

$\Rightarrow x = 4$

Number of rows

$= x = 4$

Number of students in a row

$= y = 9$

Number of total students in a class

$= x\times y = 4 \times 9 = 36$

One says, "Give me a hundred, friend! I shall then become twice as rich as you." The other replies, "If you give me ten, I shall be six times as rich as you." Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II]

Let the initial amount with them be

$\text{Rs.}\ x$ and

$\text{Rs.}\ y$ respectively.

Then according to the question:

$x + 100 = 2(y − 100)$

$x + 100 = 2y − 200$

$x − 2y = −300\dots (1)$

And,

$6(x − 10) = (y + 10)$

$6x − 60 = y + 10$

$6x − y = 70\dots (2)$

Multiplying equation

$(2)$ by 2, we get:

$12x − 2y = 140\dots (3)$

Subtracting equation

$(1)$ from equation

$(3)$ , we get:

$11x = 140 + 300$

$11x = 440$

$x = 40$

Substitute the vale of

$x$ in equation

$(1)$ , we get:

$40 − 2y = −300$

$40 + 300 = 2y$

$2y = 340$

$y =\dfrac{340}{2}= 170$

$\therefore$ The friends had

$\text{Rs.}\ 40$ and

$\text{Rs.}\ 170$ with them respectively.

In a $\triangle ABC; \angle C=3 \angle B=2(\angle A+\angle B)$ . Find the three angles.

As given that,

$\angle C = 3\angle B = 2(\angle A + \angle B)$

$3\angle B = 2(\angle A + \angle B)$

$3\angle B = 2\angle A + 2\angle B$

$\angle B = 2\angle A$

$2 \angle A − \angle B = 0$ … (i)

As the sum of the measures of all angles of a triangle is 180°. Therefore,

$\angle A + \angle B + \angle C = 180^{\circ}$

$\angle A + \angle B + 3\angle B = 180^{\circ}$

$\angle A + 4\angle B = 180^{\circ}$ ... (ii)

Multiplying equation (i) by 4, we gets

$\Rightarrow 8 \angle A − 4 \angle B = 0$ … (iii)

Adding equations (ii) and (iii), we obtain

$\Rightarrow 9 \angle A = 180^{\circ}$

$\Rightarrow \angle A = \frac{180^{\circ}}{9}=20^{\circ}$

From equation (ii), we obtain

$\Rightarrow 20^{\circ} + 4 \angle B = 180^{\circ}$

$\Rightarrow 4 \angle B = 160^{\circ}$

$\Rightarrow \angle B = 40^{\circ}$

$\Rightarrow \angle C = 3 \angle B$

$\Rightarrow 3 \times 40^{\circ} = 120^{\circ}$

$\therefore\angle A, \angle B, \angle C$ are

$20^{\circ}, 40^{\circ}, 120^{\circ}$

ABCD is a cyclic quadrilateral. Find the angles of the cyclic quadrilateral.

The sum of the opposite angles in cyclic quadrilateral is

$180^{\circ}$

$\therefore \angle A + \angle C = 180^{\circ}$

$\Rightarrow 4y + 20 − 4x = 180$

$\Rightarrow 4y-4x=180-20$

$\Rightarrow 4y-4x=160$

$\Rightarrow x-y=-40$ ....... (i)

$\angle B + \angle D = 180^{\circ}$

$\Rightarrow 3y − 5 − 7x + 5 = 180$

$\Rightarrow − 7x + 3y = 180$ ....... (ii)

Multiplying equation (i) by 3, we obtain

$\Rightarrow 3x − 3y = − 120$ ........ (iii)

Adding equations (ii) and (iii), we obtain

$\Rightarrow − 7x + 3x = 180 − 120$

$\Rightarrow − 4x = 60$

$\Rightarrow x = −15$

By using equation (i), we obtain

$\Rightarrow x − y = − 40$

$\Rightarrow −15 − y = − 40$

$\Rightarrow y = −15 + 40 = 25$

$\angle A = 4y + 20 = 4\times 25 + 20 = 120^{\circ}$

$\angle B = 3y − 5 = 3\times 25 − 5 = 70^{\circ}$

$\angle C = − 4x = − 4\times − 15 = 60^{\circ}$

$\angle D = − 7x + 5 = − 7\times −15+ 5 = 110^{\circ}$

$x+y+3z=600$
$x+ y+z=400$
In the system of equations above, what is the value of $x + y$ ?

To find the value of

$x$

$+$

$y$

Let us assume

$(x$

$+$

$y)$

$=$

$t$

Given,

$x$

$+$

$y$

$+$

$3z$

$=$

$600$

Substituting

$(x$

$+$

$y)$

$=$

$t$ , we get

$t$

$+$

$3z$

$=$

$600$

$x$

$+$

$y$

$+$

$z$

$=$

$400$

Substituting

$(x$

$+$

$y)$

$=$

$t$ , we get

$t$

$+$

$z$

$=$

$400$

Rearranging the terms, we get

$z$

$=$

$400$

$-$

$t$

From the above,

$t$

$+$

$(3$

$\times$

$z)$

$=$

$600$

Substituting

$z$

$=$

$400$

$-$

$t$ , we get

$t$

$+$

$(3$

$\times$

$(400$

$-$

$t))$

$=$

$600$

$\Rightarrow$

$t+$

$1200$

$-3t$

$=$

$600$

Rearranging the terms, we get

$2$

$\times$

$t$

$=$

$600$

$\Rightarrow t$

$=$

$\dfrac {600}{2}$

$\Rightarrow t$

$=$

$300$

$\Rightarrow t$

$=$

$(x$

$+$

$y)$

$=$

$300$

Therefore, the value of

$'(x$

$+$

$y)'$ is

$'300'$ .

Solve the following system of equations graphically. Also find the points where the lines meet the $X$ -axis.
$\displaystyle x+2y=5$
$\displaystyle 2x-3y=-4$

Given equations,

$x+2y=5$ ...(1) and

$2x-3y=-4$ ...(2)

From the graph it is clear that,point of intersection is

$(1,2)$

Point of intersection of equation(1) with X-axis =

$(5,0)$

Point of intersection of equation(2) with Y-axis=

$(-2,0)$

Solve given pair of equations, by elimination method
$3x-y=5, 5x-2y=4$

Given,

$3x-y=5$ ....(1)

And,

$5x-2y =4$ ....(2)

Multiplying Eq.1 with 2 we have:

$6x-2y=10$ ....(3)

Subtracting Eq. 2 from Eq. 3 we get:

$x=6$

And

$y=13$

Solve the following system of equations by the method of substitution:
$3a - 2b = 12$ , $4a - 5b = 16$

Given equations are

$3a-2b=12$ ....(1)

and

$4a-5b=16$ ....(2)

Therefore,

$3a=12+2b\Rightarrow a=\dfrac {12+2b}{3}$ .....(3)

Substitute this value of

$a$ in equation (2), we get

$4\left (\dfrac {12+2b}{3}\right)-5b=16$

$\Rightarrow 48+8b-15b=48$

$\Rightarrow b=0$

Put this value of

$b$ in equation (3)

$\Rightarrow a=\dfrac {12}{3}$

$\Rightarrow a=4$

Therefore, the solution is

$a=4, b=0$ .

Solve the following system of equations by the method of substitution:
$5x + 4y - 4 = 0$ , $x - 20 = 12y$

Given equations are

$5x+4y=4$ ....(1)

and

$x-20=12y$ ....(2)

Therefore,

$x=12y+20$ ....(3)

Substitute this value of

$x$ in equation (1), we get

$5(12y+20)+4y=4$

$\Rightarrow 60y+100+4y=4$

$\Rightarrow 64y=-96$

$\Rightarrow y=-\dfrac {96}{64}$

$\Rightarrow y= -\dfrac {3}{2}$

Put this value in equation (3),

$x=12\left (-\dfrac {3}{2}\right)+20$

$\Rightarrow x=-18+20=2$

Therefore, the solution is

$x=2, y=-\dfrac {3}{2}$ .

Solve the following system of equations by the method of substitution:
$3x - 4y = 10$ , $4x + 3y = 5$

Given equations are

$3x-4y=10$ ....(1)

and

$4x+3y=5$ ....(2)

Therefore,

$3x=10+4y\Rightarrow x=\dfrac {10+4y}{3}$ ....(3)

Substitute this value of

$x$ in equation (2), we get

$4\left (\dfrac {10+4y}{3}\right)+3y=5$

$\Rightarrow 40+16y+9y=15$

$\Rightarrow 25y=-25$

$\Rightarrow y=-1$

Substitute this value of

$y$ in equation (3)

$x=\dfrac {10-4}{3}=2$

Therefore, solution is

$x=2, y=-1$ .

Find the common solution:
$5x+3y=28$
$7x-2y=2$

Given equations are $5x+3y=28$ and $7x-2y=2$
Multiply $1$ st equation with $2$ and $2$ nd equation with $3$ , then add both
We get $31x = 62$ , which implies $x = 2$ and $y=(28-10)/3 = 6$

Find the common solution:
$2x+y=8$
$x-y=1$

Given $2x+y=8$ and $x-y=1$
Add above two equations we get $3x=9$ , which gives $x=3$
Therefore $y = 3-1=2$

Solve the following system of equations by the method of substitution:
$3x - 7y = 7$ , $11x + 5y = 87$

Given equations are

$3x-7y=7$ ....(1)

and

$11x+5y=87$ ....(2)

Therefore,

$3x=7+7y\Rightarrow x=\dfrac {7+7y}{3}$ ....(3)

Put this value of

$x$ in equation (2)

$11\left (\dfrac {7+7y}{3}\right)+5y=87$

$\Rightarrow 77+77y+15y=261$

$\Rightarrow 92y=261-77$

$\Rightarrow 92y=184$

$\Rightarrow y=2$

Put this value in equation (3)

Therefore,

$x=\dfrac {7+7(2)}3{}$

$\Rightarrow x=\dfrac {21}{3}$

$\Rightarrow x=7$

Therefore, solution is

$x=3, y=2$ .

Solve the following system of equations by the method of substitution:
$x + y = 10$ , $x - y = 12$

Given equations are

$x+y=10$ ....(1)

and

$x-y=12$ ....(2)

$\therefore x=12+y$ ....(3)

Substitute this value of

$x$ in equation (1), we get

$12+y+y=10$

$\Rightarrow 2y=-2$

$\Rightarrow y=-1$

Put this value in equation

$3$

Therefore,

$x=12-1$

$\Rightarrow x=11$

The solution is

$x=11, y=-1$ .

Solve the following system of equations by the method of substitution:
$2p + 3q = -5$ , $3p - 2q = 12$

Given equations are

$2p+3q=-5......(1)$

and

$3p-2q=12......(2)$

Therefore,

$3p=12+2q\Rightarrow p=\dfrac {12+2q}{3}.......(3)$

Substitute this value of

$p$ in equation

$(1)$ , we get

$2\left (\dfrac {12+2q}{3}\right) +3q=-5$

$\Rightarrow 24+4q+9q=-15$

$\Rightarrow 13q=-39$

$\Rightarrow q=-3$

Put this value of

$q$ in equation

$(3)$ .

$p=\dfrac {12+2(-3)}{3}$

$\Rightarrow p=\dfrac {12-6}{3}$

$\Rightarrow p=2$

Therefore, the solution is

$p=2, q=-3$ .

Use elimination method for solving the following equations:
$2x + y = 0$ , $3x - y = -5$

Given equations are

$2x+y=0$ ....(1)

and

$3x-y=-5$

Add both the equations to eliminate

$y$ , we get

$5x=-5$

$\Rightarrow x=-1$

substitute this value of

$x$ in equation (1),

$2(-1)+y=0$

$\Rightarrow -2+y=0$

$\Rightarrow y=2$

Therefore, the solution is

$x=-1, y=2$ .

Use elimination method for solving the following equations:
$41x + 53y = 135$ , $53x + 41y = 147$

Given equations are:

$41x+53y=135$ ....(1)

$53x+41y=147$ ....(2)

Multiply equation (1) by

$41$ and equation (2) by

$53$ we get,

$1681x+2173y=5535$ ....(3)

$2809x+2173y=7791$ ....(4)

Subtract equations (3) from equation (4) to eliminate

$y$ ,

$1128x=2256$

$\Rightarrow x=2$

Substitute this value of

$x$ in equation (1), we get

$41(2)+53y=135$

$\Rightarrow 82+53y=135$

$\Rightarrow 53y=135-82$

$\Rightarrow 53y=53$

$\Rightarrow y=1$

Therefore, the solution is

$x=2, y=1$ .

Solve the following system of equations by the method of substitution:
$2x + 3y = 5$ , $3x + 4y = 6$

Given equations are

$2x+3y=5$ ....(1)

and

$3x+4y=6$ ....(2)

Therefore,

$3x=6-4y\Rightarrow x=\dfrac {6-4y}{3}$ ....(3)

Substitute this value in equation (1), we get

$2\left (\dfrac {6-4y}{3}\right)+3y=5$

$\Rightarrow 12-8y+9y=15$

$\Rightarrow y=3$

Put this value in equation (3),

$x=\dfrac {6-12}{3}$

$\Rightarrow x=-2$

Therefore, the solution is

$x=-2, y=3$ .

Solve the following system by elimination method:
$5x - 4y = -14$ , $3x + 2y = -4$

Given equations are

$5x-4y=-14$ ....(1)

and

$3x+2y=-4$ ....(2)

Multiply equation (2) by

$2$ , we get

$6x+4y=-8$ ....(3)

Add equations (1) and (3) to eliminate

$y$ ,

$11x=-22$

$\Rightarrow x=-2$

Substitute this value of

$x$ in equation (1),

$5(-2)-4y=-14$

$\Rightarrow -10-4y=-14$

$\Rightarrow -4y=-4$

$\Rightarrow y=1$

Therefore, the solution is

$x=-2, y=1$ .

Use elimination method for solving the following equations:
$2x + y = 7$ , $2x - 3y = 3$

Given equations are

$2x+y=7$ ....(1)

and

$2x-3y=3$ ....(2)

Multiply equation (1) by

$3$ ,

$6x+3y=21$ ....(3)

Add equations (2) and (3) to eliminate

$y$ , we get

$8x=24$

$\Rightarrow x=3$

Substitute this value of

$x$ in equation (1),

$2(3)+y=7$

$\Rightarrow 6+y=7$

$\Rightarrow y=1$

Therefore, solution is

$x=3, y=1$ .

Use elimination method for solving the following equations:
$100x + 200y = 700$ , $200x + 100y = 800$

Given equations are

$100x+200y=700$ ....

$(1)$

And

$200x+100y=800$ ....

$(2)$

Multiply equation equation

$(2)$ by

$2$ , we get

$400x+200y=1600$ ....

$(3)$

Subtract equation

$(1)$ from

$(3)$ to eliminate

$y$ ,

$300x=900$

$\Rightarrow x=3$

Substitute this value of

$x$ in equation

$(1),$ we get

$100(3)+200y=700$

$\Rightarrow 300+200y=700$

$\Rightarrow 200y=400$

$\Rightarrow y=2$

Therefore, the solution is

$x=3, y=2$ .

Use elimination method for solving the following equations:
$3x + y = 10$ , $x - y = 2$

Given equations are

$3x+y=10$ ....(1)

and

$x-y=2$ ....(2)

Add both the equations to eliminate

$y$ , we get

$4x=12$

$\Rightarrow x=3$

Substitute this value of

$x$ in equation (1),

$3(3)+y=10$

$\Rightarrow 9+y=10$

$\Rightarrow y=1$

Therefore, solution is

$x=3, y=1$ .

Use elimination method for solving the following equations:
$x - y = 7$ , $3x + y = 10$

Given equations are

$x-y=7$ ....(1)

$3x+y=10$ ....(2)

Add equations (1) and (2) to eliminate

$y$ .

So, we have

$4x=17$

$\Rightarrow x=\dfrac {17}{4}$

Substitute this value of

$x$ in equation (1),

$\dfrac {17}{4}-y=7$

$\Rightarrow y=\dfrac {17}{4}-7$

$\Rightarrow y=\dfrac {17-28}{4}$

$\Rightarrow y=-\dfrac {11}{4}$

Therefore, the solution is

$x=\dfrac {17}{4}, y=-\dfrac {11}{4}$ .

Solve the following system of equations by the method of substitution:
$7x + 5y = 10$ , $3x + y = 2$

Given equations are

$7x+5y=10$ ....(1)

and

$3x+y=2$ ....(2)

Therefore,

$y=2-3x$ ....(3)

Substitute this value of

$y$ in equation (1), we get

$7x+5(2-3x)=10$

$\Rightarrow 7x+10-15x=10$

$\Rightarrow -8x=0$

$\Rightarrow x=0$

Put this value

$x$ in equation (3),

$y=2-0$

$\Rightarrow y=2$

Therefore, solution is

$x=0, y=2$ .

Solve the equations $x + y = 3$ and $3x - 2y = 4$ using graph

We draw graphs of the equations

$x + y = 3$ and

$3x - 2y = 4$ . Then locate their point of intersection.
Recall, that it is enough to know two points to draw a straight line. Consider

$x + y = 3$ . Taking

$x = 0$ , we get

$y = 3$ . Taking

$y = 0$ , we get

$x = 3$ . Thus

$(0, 3)$ and

$(3, 0)$ are on the straight line

$x + y = 3$ .
Take a graph paper. Fix your coordinate system

$x\Leftrightarrow y$ axis on the graph paper. Locate

$A = (3, 0)$ and

$B = (0, 3)$ with respect to this coordinate system.
Join

$A$ and

$B$ and extend it to a straight line.
Consider the equation

$3x - 2y = 4$ . If we take

$x = 0$ , we obtain

$y = -2$ . Similarly,

$x = 4$ gives

$y = 4$ . Thus

$C = (0, -2)$ and

$D = (4, 4)$ are points on the straight line

$3x - 2y = 4$ . Locate these points on the graph paper. Join them and extend to a straight line.
Now we have two straight line. They intersect at a point E. Looking at the graph, you see that

$E$ has coordinates

$(2, 1)$ . Thus

$x = 2$ and

$y = 1$ is the solution. We may verify it:

$2 + 1 = 3$ and

$3(2) - 2(1) = 6 - 2 = 4$ .

Solve the following system by elimination method:
$2x + 3y = 13$ , $4x + y = 11$

Given equations are

$2x+3y=13\ ....(1)$

$4x+y=11\ .....(2)$

multiply the equation

$(2)$ by

$3$ , we get

$\Rightarrow 12x+3y=33 \ .....(3)$

Subtract equation

$(1)$ from

$(3)$ to eliminate

$y$ , we get

$12x+3y=33$

$2x+3y=13$

$-$

$\overline{\ \ \ \ \ \ 10x=20\ \ \ \ \ }$

$\Rightarrow x=2$

Put this value in equation

$(1)$ , we get

$\Rightarrow2(2)+3y=13$

$\Rightarrow 4+3y=13$

$\Rightarrow 3y=9$

$\Rightarrow y=3$

$\therefore$ the solution is

$x=2, y=3$ .

Solve the following simultaneous equations using graphical method.
$x + y = 8$
$x - y = 2$

$x + y = 8$

$\therefore y = 8 - x$

$x$	$1$	$2$	$3$
$y$	$7$	$6$	$5$
$(x, y)$	$(1, 7)$	$(2, 6)$	$(3, 5)$

$x - y = 2$

$\therefore y = x - 2$

$x$	$1$	$2$	$3$
$y$	$-1$	$0$	$1$
$(x, y)$	$(1, -1)$	$(2, 0)$	$(3, 1)$

From the graph, we can clearly see that

$(x,y)=(5,3)$ is the solution to the given pair of simultaneous equations.

If $7x + 13y = 27$ and $13x + 7y = 33$ , find $x + y$

$7x + 13y = 27 ....(i)$

$13x + 7y = 33 ....(ii)$

Multiplying by

$13$ and

$7$ in equation (i) and (ii) respectively, we get

$91x + 169y = 351$

$\underline {\underset {-}{9}1x \underset {-}{+} 49y = \underset {-}{2}31}$

$120y = 120$

$y = 1$

Put the value of

$y$ in equation (i), we get

$7x + 13 \times 1 = 27$

$\Rightarrow 7x = 27 - 13$

$\Rightarrow 7x = 14$

$\Rightarrow x = 2$
Then

$x + y = 2 + 1 = 3$ .

Solve the following system by evaluating proportional value of the variables:
$3x + 2y = 4$ , $2x - 3y = 7$

Given equations are

$3x+2y=4$ ....(1)

and

$2x-3y=7$ ....(2)

Multiply equation (1) by

$3$ and equation (2) by

$2$

Therefore,

$9x+6y=12$ ....(3)

and

$4x-6y=14$ .....(4)

Now add both the equations,we get

$13x=26$

$\Rightarrow x=2$

Put this value of

$x$ in equation (1)

$3(2)+2y=4$

$\Rightarrow 6+2y=4$

$\Rightarrow 2y=-2\Rightarrow y=-1$

Therefore, the solution is

$x=2, y=-1$ .

Solve the following system by elimination method:
$3x + 2y = 5$ , $5x - 4y = 23$

Given equations are

$3x+2y=5$ ....(1)

and

$5x-4y=23$ .....(2)

Multiply equation (1) by

$2$ , we get

$6x+4y=10$ ....(3)

Now add equations both the equations to eliminate

$y$ ,

$11x=33$

$\Rightarrow x=3$

Put this value in equation (1), we get

$3(3)+2y=5$

$\Rightarrow 9+2y=5$

$\Rightarrow 2y=5-9$

$\Rightarrow y=-2$

Therefore, the solution is

$x=3, y=-2$ .

Solve the following simultaneous equations:
$\dfrac {7}{2x + 1} + \dfrac {13}{y + 2} = 27, \dfrac {13}{2x + 1} + \dfrac {7}{y + 2} = 33$

We have,

$\dfrac {7}{2x + 1} + \dfrac {13}{y + 2} = 27 ... (i)$
and

$\dfrac {13}{2x + 1} + \dfrac {7}{y + 2} = 33 .... (ii)$
Substituting

$\dfrac {1}{2x + 1} = a$ and

$\dfrac {1}{y + 2} = b$ in equations

$(i)$ and

$(ii)$ , we get

$7a + 13b = 27 ... (iii)$
and

$13a + 7b = 33 .... (iv)$
Adding equations

$(iii)$ and

$(iv)$ , we get

$20a + 20b = 60$

$a + b = 3 ... (v)$
Subtracting equation

$(iii)$ from equation

$(iv)$ , we get

$6a - 6b = 6$

$a - b = 1 ...(vi)$
Adding equations

$(v)$ and

$(vi)$ , we get

$2a = 4\Rightarrow a = 2$
Putting the value of

$a$ in equation

$(v)$ , we get

$2 + b = 3 \Rightarrow b = 1$
Resubstituting the values of

$a$ and

$b$ , we get

$\dfrac {1}{2x + 1} = 2$ and

$\dfrac {1}{y + 2} = 1$

$1 = 4x + 2$ and

$1 = y + 2$

$x = -\dfrac {1}{4}$ and

$y = -1$

Solve the following simultaneous equations using graphical method
$x + y = 6$
$x - y = 2$

$x + y = 6$

$\therefore y = 6 - x$

$x$	$0$	$1$	$2$
$y$	$6$	$5$	$4$
$(x, y)$	$(0, 6)$	$(1, 5)$	$(2, 4)$

$x - y = 2$

$\therefore y = x - 2$

$x$	$0$	$1$	$2$
$y$	$-2$	$-1$	$0$
$(x, y)$	$(0, -2)$	$(1, -1)$	$(2, 0)$

From the graph it is clear that

$(x,y) = (4,2)$ is the solution of the given pair of equations.

Solve the simultaneous equations:
$\dfrac {1}{x} + \dfrac {1}{y} = 12, \dfrac {3}{x} - \dfrac {2}{y} = 1$

$\dfrac {1}{x} + \dfrac {1}{y} = 12$

$\dfrac {3}{x} - \dfrac {2}{y} = 1$
Substituting

$\dfrac {1}{x} = a$ and

$\dfrac {1}{y} = b$ in the above equations,

$a + b = 12 ...(1)$

$3a - 2b = 1 ...(2)$
Multiplying equations

$(1)$ by

$2$ and solving with

$(2)$ , we get

$2a + 2b = 24 ...(3)$

$\underline {3a - 2b = 1}$

$5a = 25$

$a = \dfrac {25}{5} = 5$
Substituting,

$a = 5$ in equation

$(1)$ ,

$a + b = 12$

$5 + b = 12$

$b = 12 - 5 = 7$

Find ' $x+y$ ' for the following simultaneous equations : $2x+3y=12$ and $3x+2y=13$ .

Let

$2x+3y=12$ ...(1)
and

$3x+2y=13$ ...(2)
From equation (2), find the value of

$x$ in terms of

$y$

$x = \dfrac{13-2y}{3}$ ...(3)

Substituting

$x = \dfrac{13-2y}{3}$ in equation (1), we get

$\Rightarrow$

$2\left(\dfrac{13-2y}{3}\right)+3y=12$

$\Rightarrow$

$\dfrac{26-4y}{3}+3y=12$

$\Rightarrow \dfrac { 26 }{ 3 } -\dfrac { 4 }{ 3 } y+3y=12$

$\Rightarrow 3y-\dfrac { 4y }{ 3 } =12-\dfrac { 26 }{ 3 }$

$\Rightarrow \dfrac { 5y }{ 3 } =\dfrac { 10 }{ 3 }$

$\Rightarrow y=\dfrac { 10 }{ 5 } =2$

$\Rightarrow y=2$
Substituting

$y=2$ in equation (3), we get

$x=\dfrac { 13-2\times 2 }{ 3 }$

$\Rightarrow x=\dfrac { 13-4 }{ 3 }$

$\Rightarrow x=\dfrac { 9 }{ 3 } =3$

$\Rightarrow x=3$
Therefore, the value of

$x+y=3+2$ is

$5$ .

Solve the following simultaneous equations:
$12x + 13y = 62$ ;
$13x + 12y = 63$

The given simultaneous equations are,

$12x + 13y = 62$ ... (1)

$13x + 12y = 63$ .... (2)

Multiply equation (1) by

$12$ and equation (2) by

$13$ ,

$(12x + 13y)\times 12 = 62\times12$

$(13x + 12y) \times 13 = 63 \times 13$

On simplifying, we get,

$144x + 156y = 744$ .... (3)

$169x + 156y = 819$ .... (4)

Subtracting equation (3) from (4), we get

$25x = 75$

$\Rightarrow x = \dfrac {75}{25} = 3$

Substituting the value of

$x$ in equation

$(1)$ , we get

$12(3) + 13y = 62$

$\Rightarrow 36 + 13y = 62$

$\Rightarrow 13y = 62 - 36$

$\Rightarrow y= 26$

$\Rightarrow y = \dfrac {26}{13}$

$\Rightarrow y = 2$

Find ' $x+y$ ' for the following simultaneous equations : $4x+3y=8$ and $3x+4y=20$ .

Let

$4x+3y=8$ ...(1)
and

$3x+4y=20$ ....(2)
From equation (2), find the value of

$x$ in terms of

$y$ :

$x=\dfrac { 20-4y }{ 3 }$ ...(3)

Substituting

$x=\dfrac { 20-4y }{ 3 }$ in equation (1), we get

$\Rightarrow 4\left( \dfrac { 20-4y }{ 3 } \right) +3y=8$

$\Rightarrow \dfrac { 80-16y }{ 3 } +3y=8$

$\Rightarrow \dfrac { 80 }{ 3 } -\dfrac { 16y }{ 3 } +3y=8$

$\Rightarrow 3y-\dfrac { 16y }{ 3 } =8-\dfrac { 80 }{ 3 }$

$\Rightarrow \dfrac { -7 }{ 3 } y=\dfrac { -56 }{ 3 }$

$\Rightarrow y=\dfrac { 56 }{ 7 } =8$

$\Rightarrow y=8$
Substituting

$y=8$ in equation (3), we get

$x=\dfrac { 20-4\times 8 }{ 3 }$

$\Rightarrow x=\dfrac { 20-32 }{ 3 }$

$\Rightarrow x=\dfrac { -12 }{ 3 } =-4$
Therefore, the value of

$x+y$ is

$-4+8=4$ .

Find ' $x+y$ ' for the following simultaneous equations : $x+2y=2$ and $2x+y=7$ .

Let

$x+2y=2$ ...(1)
and

$2x+y=7$ ...(2)
From equation (2), find the value of

$x$ in terms of

$y$

$x = \dfrac{7-y}{2}$ ...(3)
Substituting

$x = \dfrac{7-y}{2}$ in equation (1), we get

$\Rightarrow$

$\left(\dfrac{7-y}{2}\right)+2y=2$

$\Rightarrow \dfrac { 7 }{ 2 } -\dfrac { y }{ 2 }+2y=2$

$\Rightarrow 2y-\dfrac { y }{ 2 } =2-\dfrac { 7 }{ 2 }$

$\Rightarrow \dfrac { 3y }{ 2 } =\dfrac { -3 }{ 2 }$

$\Rightarrow y=-1$
Substituting

$y=-1$ in equation (3),

$x=\dfrac { 7-\left(-1\right) }{ 2 }$

$\Rightarrow x=4$
Therefore the value of

$x+y$ is

$4+\left(-1\right)=3$ .

Solve
$\dfrac{33}{u + 2} + \dfrac{12}{v - 3} = 123$
and $\dfrac{12}{u + 2} + \dfrac{33}{v - 3} = 102$

The given simultaneous equations are

$\dfrac{33}{u + 2} + \dfrac{12}{v - 3} = 123$ ........ (i)

$\dfrac{12}{u + 2} + \dfrac{33}{v - 3} = 102$ ........ (ii)
Substituting

$\dfrac{1}{u + 2} = m$ and

$\dfrac{1}{v - 3} = n$ then adding, we get

$33 m + 12 n = 123$ ....... (iii)

$12 m + 33 n = 102$ ....... (iv)
----------------------------------

$45 m + 45 n = 225$

$45(m + n) = 225$

$m + n = 5$ ........ (v)
Substracting equation (iv) from (iii), we get

$33 m + 12 n = 123$

$12 m + 33 n = 102$
- (-) (-)
-------------------------------

$21 m - 21 n = 21$

$21(m - n) = 21$

$m - n = 1$ ....... (vi)
Now, adding equation (v) and (vi), we get

$m + n = 5$

$m - n = 1$
--------------------

$2m = 6$

$m = 3$
Substituting m = 3 in equation (v), we get

$m + n = 5$

$3 + n = 5$

$n = 2$
Re-substituting values of m and n.

$\dfrac{1}{u + 2} = 3$

$\dfrac{1}{v - 3} = 2$

$3u + 6 = 1$

$2v - 6 = 1$

$3u = - 5$

$2v = 1 + 6$

$2v = 7$

$u = \dfrac{-5}{3}$

$v = \dfrac{7}{2}$

Solve the following simultaneous equations:
$\cfrac { 10 }{ x+y } +\cfrac { 2 }{ x-y } =4$ ;
$\cfrac { 5 }{ x+y } -\cfrac { 5 }{ 3(x-y) } =\cfrac { -2 }{ 3 }$

Given,

$\cfrac { 10 }{ x+y } +\cfrac { 2 }{ x-y } =4$

$\cfrac { 5 }{ x+y } -\cfrac { 5 }{ 3(x-y) } =-\cfrac { 2 }{ 3 }$

Substituting

$\cfrac{1}{x+y}=a$ and

$\cfrac{1}{x-y}=b$ in the above equations

$\therefore$

$10a+2b=4........(1)$

$15a-5b=-2..........(2)$

Multiplying equations

$(1)$ by

$5$ and

$(2)$ by

$2$

$5(10a+2b=4)$

$50a+10b=20........(3)$

$2(15a-5b=-2)$

$30a-10b=-4..........(4)$

Adding equation

$(3)$ and

$(4)$

$50a+10b=20$

$30a-10b=-4$
----------------------

$80a=16$

$a=\cfrac{16}{80}$

$\therefore$

$a=\cfrac{1}{5}$
Substituting

$a=\cfrac{1}{5}$ on equation

$(1)$

$10a+2b=4$

$10\times \cfrac{1}{5}+2b=4$

$2+2b=4$

$2b=4-2$

$2b=2$

$b=1$

$\cfrac{1}{x+y}=\cfrac{1}{5}$ ,

$\cfrac{1}{x-y}=1$

$x+y=5.........(5)$

$x-y=1..........(6)$
-------------------

$2x=6$ .......... (Adding)

$x=\cfrac{6}{2}$

$x=3$

Substituting

$x=3$ in equation

$(5)$ we get

$x+y=5$

$3+y=5$

$y=5-3$

$\therefore$

$y=2$
Solution set:

$x=3; y=2$

Find ' $m+n$ ' for the following simultaneous equations : $7m+3n=12$ and $3m+7n=8$ .

Let

$7m+3n=12$ ...(1)
and

$3m+7n=8$ ...(2)
From equation (2), find the value of

$m$ in terms of

$n$

$m = \dfrac{8-7n}{3}$ ...(3)
Substituting

$m = \dfrac{8-7n}{3}$ in equation (1)

$\Rightarrow$

$7\left(\dfrac{8-7n}{3}\right)+3n=12$

$\Rightarrow \dfrac { 56 }{ 3 } -\dfrac { 49n }{ 3 }+3n=12$

$\Rightarrow \dfrac { -49n }{ 3 }+3n =12-\dfrac { 56 }{ 3 }$

$\Rightarrow \dfrac { -40n }{ 3 } =\dfrac { -20 }{ 3 }$

$\Rightarrow n=\dfrac { -20 }{ -40 }$

$\Rightarrow n=\dfrac{1}{2}$
Substituting

$n=\dfrac{1}{2}$ in equation (3), we get

$m=\dfrac { 8-7\left(\dfrac{1}{2}\right) }{ 3 }$

$\Rightarrow m=\dfrac { \dfrac{9}{2} }{ 3 }$

$\Rightarrow m=\dfrac { 9 }{ 6 }$

$\Rightarrow m=\dfrac{3}{2}$
Therefore, the value of

$m+n$ is

$\dfrac{3}{2}+\dfrac{1}{2}=\dfrac{4}{2}=2$ .

The cost of 2 exercise books and 3 pencils is Rs. 17 and the cost of 3 exercise books and 4 pencils is Rs.Formulate the problem algebrically and solve it graphically.

Let the cost of an exercise book be

$Rs. x$

Let the cost of a pencil be

$Rs. y$

$2x+3y = 17$ Blue line

$3x+4y=24$ Red line

Corresponding graph is shown in the attached figure. From the graph, it can be seen the solution is

$x=4,y=3$ .

Hence, the cost of exercise book is

$Rs. 4$ and that of a pencil is

$Rs. 3$ .

Solve the following simultaneous equations using graphical method
$4x = y - 5$
$y = 2x + 1$

$4x=y-5$

$x$	$0$	$2$	$-2$
$y$	$5$	$13$	$-3$

$y = 2x+1$

$x$	$0$	$2$	$-2$
$y$	$1$	$5$	$-3$

Plotting both the equations on graph, as shown,

$(x,y) = (-2,-3)$ is the solution of the given pair of equations.

Solve the following pair of equations by reducing them to a pair of linear equations
$\cfrac { 5 }{ x-1 } +\cfrac { 1 }{ y-2 } =2$ and $\cfrac { 6 }{ x-1 } -\cfrac { 3 }{ y-2 } =1$

First equation:

$\dfrac{5}{x-1} + \dfrac{1}{y-2} =2$
or,

$\dfrac{5(y-2)+x-1}{(x-1)(y-2)} = 2$

$\dfrac{5y-10+x-1}{xy-2x-y+2} = 2$
=

$x+5y-11 = 2xy -4x-2y+4$
or ,

$x+5y-11+4x+2y-4 = 2xy$
or,

$5x+7y-15 =2xy$
second equation:

$\dfrac{6}{x-1} - \dfrac{3}{y-2} =1$

$\dfrac{6y-12-3x+3}{xy-2x-y+2} = 1$
=

$6y-3x-9 = xy -2x-y+2$

$6y-3x-9+2x+y-2 = xy$
or,

$7y-x-11 = xy$
Multiply the second equation by 2 and subtract first equation from resultant

$14y-2x-22 =2xy$

$7y+5x-15 = 2xy$
------------------------------

$7y-7x-7 = 0$

$y-x-1 = 0$

$y = x+1$
Substituting the value of x in first equation, we get;

$5x + 7y - 15 = 2xy$
Or,

$5x + 7x + 7 - 15 - 2x(x + 1)$ Or,

$12x- 8 = 2x^2 + 2x$ Or,

$10x - 8 = 2x^2$ Or,

$x^2 = 5x- 4$
Similarly, substituting the value of

$y$ in second equation, we get;

$7y - x - 11 = xy$
Or,

$7x + 7 - x - 11 = x^2 + 1$ Or,

$6x- 5 = x^2$
From above two equations, it is clear

$5x - 4 = 6x - 5$
Or,

$5x + 1 = 6x$
Or,

$x = 1$
Hence,

$y = x + 1 = 2$
Hence,

$x = 1$ and

$y = 2$

Solve the pair of linear equations graphically.
$2x + y - 6 = 0$
$4x - 2y - 4 = 0$

clearly, from graph

$x=2,y=2$ is solution. (intersection point)

Draw a graph for the following pair of linear equations in two variables and find their solution from the graph.
$2x+y=5$
$3x-2y=4$

Clearly, from figure graph of both lines cut at

$x=2, y=1$
Hence, solution is

$x=2,y=1$

Solve the equations
$\dfrac{5}{x - 1} + \dfrac{1}{y - 2} = 2$ and $\dfrac{6}{x - 1} - \dfrac{3}{y - 2} = 1$

Let

$\dfrac 1{x-1} = X$ and

$\dfrac 1{y-2} = Y$

Hence, the given equations are reduced to:

$5X + Y = 2$ ....

$(1)$

$6X - 3Y = 1$ ....

$(2)$

Multiplying

$(1)$ by

$3$ , we get

$15X + 3Y = 6$ ....

$(3)$

Adding

$(3)$ and

$(2)$ , we get

$21X = 7$

$\Rightarrow X = \dfrac13$

Substitute this value of

$X$ in

$(2)$ , we get

$6\left(\dfrac13\right) - 3Y = 1$

$\Rightarrow 2-3Y =1$

$\Rightarrow 3Y =1$

$\Rightarrow Y = \dfrac13$

Resubstitute the values of

$X$ and

$Y$ to get

$x$ and

$y$ ,

$X = \cfrac1{x-1}$

$\Rightarrow \cfrac 13 = \cfrac1{x-1}$

$\Rightarrow x-1 = 3$

$\Rightarrow x = 4$

$Y = \cfrac1{y-2}$

$\Rightarrow \cfrac13 = \cfrac1{y-2}$

$\Rightarrow y -2 = 3$

$\Rightarrow y = 5$

Hence,

$x = 4, y=5$ is the solution of the given simultaneous equations.

Draw the graph of $2x+3y=12$ .

First, we rewrite the equation

$2x+3y=12$ in the form of

$y=mx+c$ .

$2x+3y=12$ implies

$y=-\dfrac{2}{3}x+4$
Substituting

$x=-3, 0, 3$ in the above equation, we find the values of

$y$ as follows

$y=-\dfrac{2}{3}x+4$

$x$	$-3$	$0$	$3$
$-\dfrac{2}{3}x$	$2$	$0$	$-2$
$y=-\dfrac{2}{3}x+4$	$6$	$4$	$2$

Plot the points

$\left( -3,6 \right) ,\left( 0,4 \right)$ and

$\left( 3,2 \right)$ and draw a line passing through these points. Now we get the required graph.
1028908_618904_ans_8b768d3583a844b9b0c52e787edd65e9.bmp

1028908_618904_ans_8b768d3583a844b9b0c52e787edd65e9.bmp

Solve the following pair of linear equations by Substitution method
$2x-3y=19$ and $3x-2y=21$ .

The equations given to us are:

$2x - 3y = 19$ ...

$(1)$

$3x - 2y = 21$ ...

$(2)$

By substitution method, we will substitute the value of

$y$ from equation

$1$ in equation

$2$ .

$\therefore$ from first equation

$y = \dfrac{(2x - 19)}{3}$

So by substituting the value of

$y$ in the second equation,

The second equation becomes:

$3x-\dfrac{2(2x-19)}{3}=21$

$\Rightarrow \dfrac{9x-4x+38}{3}=21$

$\Rightarrow 5x=63-38$

$\Rightarrow 5x=25$

$\Rightarrow x = 5$

Substituting the value of

$x$ in equation number

$(1)$ we get ,

$2(5) - 3y = 19$

$\Rightarrow 10-19 = 3y$

$\Rightarrow -9 = 3y$

$\Rightarrow y = -3$

then

$y$ from above equation will be

$-3$ .

The coach of a cricket team buys one bat and $2$ balls for $Rs. 300$ . Later he buys another $2$ bats and $3$ balls of the same kind for $Rs. 525$ . Represent this situation algebraically and solve it by graphical method. Also, find out that how much money the coach will pay for the purchase of one bat and one ball.

Let cost of one bat be

$Rs.\ x$

Let cost of one ball be

$Rs.\ y$

$x+2y = 300 ...........(i)$ Blue line

$2x+3y = 525 ...........(ii)$ Red line

The graphs for the equation are shown in the attached plot.

From graph, solution is

$x =150, y=75$

Hence, cost of one bat is

$Rs.\ 150$ and cost of one ball is

$Rs.\ 75$

Cost of one bat and one ball is

$Rs.\ 150+Rs.\ 75 = Rs.\ 225$

Solve the following equations graphically.
$3x - y = 7$ ,
$2x + 3y = 1$

Clearly, from figure both line intersect at

$x=2,y=-1$
Therefore, solution is

$x=2,y=-1$

Solve by substitution method $\dfrac{1}{x} + \dfrac{1}{y} = 4$ and $\dfrac{2}{x} + \dfrac{3}{y} = 7, x \neq 0, y \neq 0$

Let

$\dfrac{1}{x} = a$ and

$\dfrac{1}{y} = b$
The given equations become

$a + b = 4$ ........ (1)

$2a + 3b = 7$ ......... (2)
Equation (1) becomes b = 4 - a ......... (3)
Substituting b in (2) we get,

$2a + 3 (4 - a) = 7$

$\Rightarrow 2a + 12 - 3a = 7$

$2a - 3a = 7 - 12$

$- a = - 5 \Rightarrow a = 5$
Substituting a = 5 in (3) we get,

$b = 4 - 5 = - 1$
But

$\dfrac{1}{x} = a \Rightarrow x = \dfrac{1}{a} = \dfrac{1}{5}$

$\dfrac{1}{t} = b \Rightarrow y = \dfrac{1}{b} = \dfrac{1}{-1} = - 1$

$\therefore$ The solution is

$x = \dfrac{1}{5} , y = - 1$

Solve $x + 3y = 16, 2x - y = 4$ by using substitution method.

We have

$x + 3y = 16$ ..... (1)

$2x - y = 4$ ...... (2)

Equation (1) becomes, x = 16 - 3y ...... (3)

Substituting the above relation of

$x$ in (2) and we got,

$\Rightarrow 2(16 - 3y) - y = 4$

$\Rightarrow 32 - 6y - y = 4$

$\Rightarrow - 6y - y = 4 - 32$

$\Rightarrow -7y = - 28$

$\Rightarrow y = \dfrac{-28}{-7} = 4$

Substituting

$y = 4$ in eq. (3) we got,

$x = 16 - 3(4)$

$= 16 - 12 = 4$

$\therefore$ The solution is x = 4 and y = 4.

Solve by elimination method $3x+4y=-45, 2x-3y=6$ .

The given system is

$3x+4y=-25$ (1)

$2x-3y=6$ (2)
To eliminate the variable x, let us multiply (1) by 2 and (2) by -3 to obtain

$(1) \times 2 \rightarrow 6x+8y=-50$ (3)

$(2) \times -3 \rightarrow -6x+9y=-18$ (4)
Now, adding (3) and (4) we get, y 17 = 68 which gives y = 4 Next, substitute y y = - 4 in (1) to obtain

$3x+(-4) =-25$
That is,

$x=-3$
Hence, the solution is ( -3, -4 ).

Solve $3x- 5y = -16, 2x+ 5y = 31$

The given equations are

$3x- 5y = -16$ (1)

$2x+ 5y = 31$ (2)
Note that the coefficients of y in both equations are numerically equal. So, we can eliminate y easily.
Adding (1) and (2), we obtain an equation

$5x=15$
That is,

$x=3$ (3)
Now, we substitute x = 3 in (1) or (2) to solve for y.
Substituting x = 3 in (1) we obtain,

$3(3) -5y = -16$

$\rightarrow y=5$
Now, (3, 5) a is solution to the given system because (1) and (2) are true when x = 3 and y = 5 as from (1) and (2) we get,

$3(3) - 5(5) = -16$ and

$2(3) +5(5) = 31$ .

Prove that there is a value of $c$ for which the system-
$cx + 2y = c - 2$
$8x + cy = c$
has infinitely many solutions. Find this value.

Given equations are

$cx + 2y = c - 2$

$8x + cy = c$
Here

$a_{1} = c, b_{1} = 2, c_{1} = c - 2; a_{2} = 8, b_{2} = c, c_{2} = c$

When the system of equations has infinite solutions, then

$\dfrac {a_{1}}{a_{2}} = \dfrac {b_{1}}{b_{2}} = \dfrac {c_{1}}{c_{2}}$

$\therefore \dfrac {c}{8} = \dfrac {2}{c} = \dfrac {c - 2}{c}$

From first two terms,

$\dfrac {c}{8} = \dfrac {2}{c}$

$\Rightarrow c^{2} = 16$

$\Rightarrow c = \pm 4$

From last two terms,

$\dfrac {2}{c} = \dfrac {c - 2}{c}$

$\Rightarrow 2 = c - 2$

$\Rightarrow c = 2 + 2 = 4$
So, the required solutions are

$\ c =\pm 4$

Find the solution of given pair of linear equation by elimination method
$3x + 4y = - 17$ ...... (1)
$5x + 2y = - 19$ ....... (2)

$3x + 4y = - 17$ ........ (1)

$5x + 2y = - 19$ ........ (2)
Multiplying equation (2) by 2, we get

$10 x + 4y = - 38$ ....... (3)
Subtracting equation (3) from equation (1), we have

$- 7x = 21$

$\Rightarrow x = - 3$
Substituting

$x = -3$ in equation (1), we get

$3(-3) + 4y = - 17$

$\Rightarrow -9 + 4y = - 17$

$\Rightarrow 4y = - 17 + 9 = - 8$

$\Rightarrow y = - \dfrac{8}{4} = - 2$

$\therefore$

$(x, y) = (-3, -2)$ is the solution of the given pair of linear equations.

Solve the following system of equations by elimination method:
$3x+5y=20$
$6x-10y=-4$

Given system of equations

$3x+5y=20........(i)$

$6x-10y=-4.......(ii)$

Multiply both sides by

$2$ in equation

$(i)$

$6x+10y=40.......(iii)$

$6x-10y=-4........(ii)$

By addition

$(6x+10y)+(6x-10y)=40-4$

$\Rightarrow$

$12x=36$

$\Rightarrow$

$x=\cfrac{36}{12}$

$\Rightarrow$

$x=3.......(iv)$
Putting the value of

$x$ from

$(iv)$ in equation

$(i)$

$\Rightarrow$

$3\times 3+5y=20$

$\Rightarrow$

$9+5y=20$

$\Rightarrow$

$5y=20-9$

$\Rightarrow$

$5y=11$

$\Rightarrow$

$y=\cfrac{11}{5}$

Hence, the solution is

$x=3,y=\cfrac{11}{5}$

Solve the following system of equation by Substitution method.
$x+y=7$
$3x-2y=11$

Given the system of equation

$x+y=7.......(i)$

$3x-2y=11....(ii)$
From equation

$(i)$

$x+y=7$

$x=7-y.......(iii)$
Putting the value of

$x$ from

$(iii)$ in equation

$(ii)$

$3(7-y)-2y=11$

$\Rightarrow$

$21-3y-2y=11$

$\Rightarrow$

$21-5y=11$

$\Rightarrow$

$-5y=11-21$

$\Rightarrow$

$y=\cfrac{10}{5}=2$

$\Rightarrow$

$y=2........(iv)$
Putting the value of

$y$ from

$(iv)$ in equation

$(iii)$

$\Rightarrow$

$x=7-2$

$\Rightarrow$

$x=5$

$\Rightarrow$

$x=5$ and

$y=2$

Prove that there is a value of c for which the system $cx+2y=c-2, 8x+cy=c$ has infinitely many solutions. Find this value.

Two equations given are

$cx + 2y = c-2$

$8x + cy =c$

For infinite solutions

$\dfrac { { a }_{ 1 } }{ { a }_{ 2 } } =\dfrac { { b }_{ 1 } }{ { b }_{ 2 } } =\dfrac { { c }_{ 1 } }{ { c }_{ 2 } }$

$\dfrac { c }{ 8 } =\dfrac { 2 }{ c } =\dfrac { { c }-2 }{ { c } }$

Now

$\dfrac { c }{ 8 } =\dfrac { 2 }{ c }$

${ c }^{ 2 }=16$

$c=\pm 4$

now

$\dfrac { 2 }{ c } =\dfrac { { c }-2 }{ { c } }$

${ c }^{ 2 }-4c=0$

$c(c-4)=0$

$c = 0,4$

so only common value of c is

$4$

so for

$c =4$ this system equations posses the infinite solutions.

Solve the following system of equations of elimination by equating the coefficients method: $3x-4y-11=0, 5x-7y+4=0$ .

Given

$3x-4y-11=0 ....(1)$

$5x-7y+4=0 ...(2)$

Multiply equations (1) and (2) by

$5$ and

$3$ respectively

We get

$15x-20y-55=0$ ----(1) and

$15x-21y+12=0$ .........(ii)

subtract the eq(i) from eq(ii) we get the below equation,

$-20y-55=-21y+12$

On solving we get

$y=67$

$x=93$

Solve the following system of equations by elimination method.
$3x - 4y - 11 = 0$
$5x - 7y + 4 = 0$ .

$3x - 4y - 11 = 0 ... (i)$

$5x - 7y + 4 = 0 ... (ii)$
Multiply (i) by

$7$ and (ii) by

$4$ , then subtract (ii) from (i)

$21x - 28y - 77 = 0$

$\underline {\underset {-}{2}0x \underset {+}{-}28y \underset {-}{+} 16 = 0}$

$x - 93 = 0$

$x = 93$
Put the value of

$x$ in equation (i)

$3x - 4y - 11 = 0$

$3\times 93 - 4y - 11 = 0$

$279 - 4y - 11 = 0$

$268 - 4y = 0$

$-4y = -268$

$y = \cfrac {-268}{-4}$

$y = 67$

Hence,

$x = 93$ and

$y = 67$ .

Solve the following system of equation by elimination method:
$3x+2y=11,2x+3y=4$

$3x+2y=11........(1)$

$2x+3y=4..........(2)$
Multiplying equation

$(1)$ by

$2$ and equation

$(2)$ by

$3$ and subtracting

$\left( {6x + 4y} \right) - \left( {6x + 9y} \right) = 22 - 12$

$\Rightarrow - 5y = 10$

$\Rightarrow y = - 2$

Substitute the value of

$y$ in equation

$(1),$

$3x+2\times (-2)=11$

$\Rightarrow$

$3x=11+4=15$

$\Rightarrow$

$x=\cfrac{15}{3}=5$

$\Rightarrow x=5$

$\therefore$ Solution of system of equations is

$x=5,y=-2.$

Solve the following systems of equations graphically:
$x-2y=5$
$2x+3y=10$

We can see that two lines intersect at the point

$x=5,y=0$ . So, the solution of the given system of equation is

$(5,0)$ .

Solve graphically the system of linear equations:
$4x-3y+4=0$
$4x+3y-20=0$
Find the area bounded by these lines and $x$ -axis.

$\dfrac { x }{ 7 } +\dfrac { y }{ \left( 4/3 \right) } =1\quad \longrightarrow \left( a \right)$

$\dfrac { x }{ 5 } +\dfrac { y }{ \left( 20/3 \right) } =1\quad \longrightarrow \left( b \right)$

Intersection

$\rightarrow$

$x=2$ ,

$y=3$

Area

$=\dfrac { 1 }{ 2 } \times 6\times 3=9$ sq. units.

1225350_966005_ans_a53a75c30be345c583554303deb5155a.png

Solve the following system of linear equations graphically:
$4x-5y-20=0$
$3x+5y-15=0$
Determine the vertices of the triangle formed by the lines representing the above equation and the $y$ -axis.

Taking equation

$(1)$

$4x-5y-20=0$

$x$

$0$

$5$

$Y$

$-4$

$0$

tAKING EQUATION

$(2)$

$3X+5Y-15=0$

$X$

$0$

$5$

$Y$

$3$

$0$

$\Rightarrow \$ Solution is given by the point where

$2$ times intersect each others i.e

$(5,0)$

$\therefore \ x=5\ y=0$ (required solution)

$\Rightarrow$ Vertices of triangle

$ABC=(0,3)(5,0)(0,-4)$

1441400_966025_ans_ac2d13cb4d79492da3145febf90d6ab1.png

Solve the following system of equations graphically:
$2x-3y+6=0$
$2x+3y-18=0$

${\textbf{Hint: Draw both lines on graph and find point of intersection}}$

${\textbf{Step - 1: Finding table of value satisfying the given equation}}$

${\text{Given, }}{{\text{L}}_{\text{1}}} \equiv {\text{ 2x }}{\text{ - 3y + 6 = 0}}$

${\text{and }}{{\text{L}}_{\text{2}}}{\text{ }} \equiv {\text{ 2x + 3y - 18 = 0}}$

${\text{For }}{{\text{L}}_{\text{1}}}{\text{ }}\equiv 2x-3y+6=0$

$x$	0	3
$y=\dfrac{2x+6}{3}$	2	4

${\text{For }}{{\text{L}}_{\text{2}}}{\text{}}\equiv 2x+3y-18=0$

$x$	0	9
$y=\dfrac{-2x+18}{3}$	6	0

$\textbf{Step 2: Plot these points on XY-Plane and join them to get the graph of the given lines.}$

${\textbf{Step - 3: Finding intersection point}}$

${\text{From the graph, we can clearly see that the two lines are intersecting at a point C(3,4)}}{\text{.}}$

$\textbf{Hence, the solution of given pairs of equation is }$

$\boldsymbol{x=3, y=4}$

Solve the following systems of equations graphically:
$x+y=3$
$2x+5y=12$

We can see in the figure that two lines intersect at a point in 1st quadrant

Multiply 1st equation on both sides by 2

$\Rightarrow 2x+2 y=6$

$\Rightarrow 2x+5y=12$

Subtract two equations ;

$\Rightarrow 3y=6$

$\Rightarrow y=2$

$\Rightarrow x=3-y=3-2=1$

$\Rightarrow x=1,y=2$

Draw the graphs of $x-y+1=0$ and $3x+2y-12=0$ . Determine the coordinates of the vertices of the triangle formed by these lines and $x$ -axis and shade the triangular area.

Given,

$x-y+1=0$ ....(1)

$3x+2y-12=0$ ....(2)

put

$x=0$ in (1)

$0-y+1=0\Rightarrow y=1$

$(0,1)$ .....(a)

put

$y=0$ in (2)

$3x+2(0)-12=0\rightarrow x=4$

$(4,0)$ .....(b)

put

$x=0$ in (2)

$3(0)+2y-12=0\rightarrow y=6$

$(0,6)$ ......(c)

put

$y=0$ in (1)

$x-0+1=0\Rightarrow x=-1$

$(-1,0)$ ...(d)

From figure, the vertices of triangle are

$(-1,0),(2,3),(4,0)$

1433866_966004_ans_4260223fca9d4f269f7e3551126f342d.png

Solve the following system of linear equations graphically:
$3x+y-11=0,\ x-y-1=0.$
Shade the region bounded by these lines and $y$ -axis. Also, find the area of the region bounded by the these lines and $y$ -axis.

Given,

$3x+y-11=0$

$\Rightarrow 3x+y=11$ ........(1)

$x-y-1=0$

$\Rightarrow y=x-1$ ......(2)

substitute the value of "y" in (1), we get,

$3x+(x-1)=11$

$\therefore x=3$

$\therefore y=2$

$(x,y)=(3,2)$

area of triangle formed by vertices

$a(3,2),b(0,11),c(0,-1)$

$A=\dfrac{1}{2}\times BC\times AD$

$=\dfrac{1}{2}\times 12\times 3$

$=18sq. units$

1408528_966007_ans_8c057678e8c64dddb9b3f0804ff40af4.png

Represent the following pair of equations graphically and write the coordinates of points where the line intersects $y$ -axis
$x+3y=6$
$2x-3y=12$

$x+3y=6$

$2x-3y=12$

plotting graph

$:$

$x+3y=6$

$x$	$0$	$6$	$3$
$y$	$2$	$0$	$-1$

plotting graph

$:$

$2x-3y=12$

$x$	$0$	$6$	$9$
$y$	$-4$	$0$	$2$

$y\,$ axis coordinates are (0,2) and (0,-4).

1005511_966521_ans_0032e5c9502244dbb44a88ac011ccd09.png

Show graphically that each one of the following systems of equations is in-consistent (i.e. has no solution):
$3x-5y=20$
$6x-10y=-40$

We can see that both the lines have same Slope=

$\dfrac {3}{5}$

Constant of two lines is different

$\Rightarrow$ lines are parallel

$\therefore$ lines have no solution ie, they do not intersect and system of equations is in-consistent

Meena went to a bank to withdraw $Rs\ 2000$ . She asked the cashier to give her $Rs\ 50$ and $Rs\ 100$ notes only. Meena got $25$ notes in all. Find how many notes $Rs\ 50$ and $Rs\ 100$ she received.

Solve graphically the following systems of linear equations. Also find the coordinates of the points where the lines meet axis of $y$ .
$x+2y-7=0$ ,
$2x-y-4=0$

The lines

$:$

$x+2y-7=0$

$2x-y-4=0$

plotting graph

$:$

$x+2y-7=0$

$x$	$0$	$7$	$-1$
$y$	$\cfrac {7}{2}$	$0$	$4$

plotting graph

$:$

$2x-y-4=0$

$x$	$0$	$2$	$3$
$y$	$-4$	$0$	$2$

The solution

$(3,2)$

The point where they meet

$y$ -axis

$,$

$(0,\cfrac {7}{2})$ and

$(0,-4)$

$.$

1005340_987867_ans_6ba4c62652a140d1aa5861741386a6eb.png

$12x + 10y = 13000$
$10x + 12y = 13400$
Find $x,y$

$12x+10y=13000\longrightarrow(i)$

$10x+12y=13400\longrightarrow(ii)$

Multiplying

$(i)$ by

$5$ and

$(ii)$ by

$6$ , we set ,

$60x+50y=65000\longrightarrow(i)$

$60x+72y=80400\longrightarrow(ii)$

On substraction :

$\Rightarrow -22y=-15400$

$\therefore y=700$

Putting the value of

$y$ in equation

$\longrightarrow(i)$

$12x+(700)(10)=13000$

$\Rightarrow 12x=6000$

$\therefore x=500$

$y=700$

Solve $3 - \left( {x - 5} \right) = y + 2$
$2\left( {x + y} \right) = 4 - 3y$

$3-\left( x-5 \right) =y+2\\ \Rightarrow 3-x+5=y+2\\ \Rightarrow x+y=6\longrightarrow (i)$

And

$,$

$2\left( x+y \right) =4-3y\\ \Rightarrow 2x+2y=4-3y\\ \Rightarrow 2x+5y=4\longrightarrow (ii)$

Multiplying

$(i)$ by

$2$

$2x+2y=12\longrightarrow(iii)$

On substraction

$(ii)$ and

$(iii)$

$:$

$3y=-8$

$\Rightarrow y=-\cfrac{8}{3}$

Putting value of

$y$ in equation

$(i)$

$x-\cfrac{8}{3}=6$

$\therefore x=\cfrac{26}{3}$

Solve the following system of equations by using the method of substitution,
$3x-5y=-1$ and $x-y=-1$ .

The given system of equations is

$3x-5y=-1$ ..... (i)

$x-y=-1$ ......(ii)

From (ii), we get

$y=x+1$

Substituting

$y=x+1$ in (i), we get

$3x-5(x+1)-1$

$\Rightarrow -2x-5=-1$

$\Rightarrow -2x=4\Rightarrow x=-2$

Putting

$x=-2$ in

$y=x+1$ we get

$y=-1$ .

Hence, the solution of the given system of equations is

$x=-2, y=-1$ .

Find $x+y$ for the following system of linear equations graphically,
$x-y=1$ and $2x+y=8$ .

Given linear equation are

$x-y=1$ and

$2x+y=8$

let us find two points on the line

$x-y=1$

$x-y=1\Rightarrow x-1=y$

For

$x=1\Rightarrow y=0$

For

$x=2\Rightarrow y=1$

Now we find two points on the line

$2x+y=8$

$2x+y=8\Rightarrow y=8-2x$

For

$x=1\Rightarrow y=6$

For

$x=2\Rightarrow y=4$

Lets us plot those line on graph

So we see lines intersect at point

$(3,2)$ . Hence

$x=3, y=2$

Hence

$(x+y)=(3+2)$

$=5$

1435037_1008011_ans_c2757496e74b495d9ff2b5c2265a2749.png

Solve: $2x - 3y - 3 = 0$
$\dfrac{{2x}}{3} + 4y + \dfrac{1}{2} = 0$

$2x-3y-3=0\longrightarrow (i)\\ \cfrac { 2x }{ 3 } +4y+\cfrac { 1 }{ 2 } =0\\ \Rightarrow \cfrac { 4x+24y+3 }{ 6 } =0\\ 4x+24y+3=0\longrightarrow (ii)\\$

Multiplying

$(i)$ by

$2.$

$\quad\quad\quad\quad\quad\quad4x-6y-6=0\\ \quad\quad\quad\quad\quad\quad4x+24y+3=0$

On substraction :

$-30y-9=0\\ y=\cfrac { 9 }{ 30 } \\ =\cfrac { 3 }{ 10 }$

Putting value of

$y$ in equation

$\longrightarrow(i)$

$2x-3\left( \cfrac { 3 }{ 10 } \right) -3=0\\ \Rightarrow 2x=\cfrac { 9 }{ 10 } +3\\ \Rightarrow 2x=\cfrac { 39 }{ 10 } \\ \Rightarrow x=\cfrac { 39 }{ 20 } \\ x=\cfrac { 39 }{ 20 } \\ y=\cfrac { 3 }{ 10 } .$

Solve the following system of equations graphically
$x+3y=6$
$2x-3y=12$
and hence find the value of a, if $4x+3y=a$ .

1580996_1008004_ans_45d614608fe64ac1a9f5c20e615a97b9.jpg

Solve the following system of linear equations graphically. Also, find the coordinates of the points where the lines meet the $y\ -$ axis.

$2x-y-5=0$
$x-y-3=0$

Given lines:

$2x-y-5=0$ and

$x-y-3=0$

The line

$2x-y-5=0$ can be written as

$y=2x-5$

$...(1)$

$x$	0	1	3
$y$	-5	-3	1

The line

$x-y-3=0$ can be written as

$x=3+y$

$...(2)$

$x$	0	3	4
$y$	-3	0	1

Let us substitute

$x=3+y$ in equation

$(1)$

$\Rightarrow 2(3+y)-y-5=0$

$\Rightarrow 6+2y-y-5=0$

$\Rightarrow y=-1$

$\therefore\ x=2$

$\therefore$ The lines meet at the point

$(2,-1)$

The coordinates where the lines meet

$y\ -$ axis are

$(0,-3)$ and

$(0,-5).$

1005296_987876_ans_368bd86f2bc943a0afaf68fa1215c869.png

Solve graphically the following systems of linear equations. Also find the coordinates of the points where the lines meet axis of $y$ .
$2x+y-11=0$ ,
$x-y-1=0$

$2x+y-11=0$

$x-y-1=0$

plotting graph

$:$

$2x+y-11=0$

$x$

$0$

$\cfrac {11}{2}$

$5$

$y$

$11$

$0$

$1$

plotting graph

$:$

$x-y-1=0$

$x$	$0$	$1$	$2$
$y$	$-1$	$0$	$1$

The solution is

$(4,3)$

The points where the lines meet

$y$ -axis are

$:$

$(0,11)$ and

$(0,-1)$

$.$

1005265_987861_ans_b9037521fdf641788d1330486a6919e6.png

Determine, by drawing graphs, whether the following system of linear equation has a unique solution or not:
$2y=4x-6,\ 2x=y+3$

Given

$2y=4x-6$

$\implies 2x-y-3=0----(1)$

$2x=y+3\implies 2x-y-3=0-----(2)$

So from graph there exists infinite number of solutions.

Solve the following system of equations:
$8v=3u+5uv$
$6v-5u=-2uv$ .

Clearly, the given equations are not linear in the variables

$u$ and

$v$ but can be reduced into linear equations by appropriate substitution.

If we put

$u=0$ in either of the two equations, we get

$v=0$

Thus,

$u=0, v=0$ from one solution of the given system of equations.

To find the other solutions, we assume that

$u\neq 0$ ,

$v\neq 0$ .

Since

$u\neq 0$ ,

$v\neq 0$ . Therefore,

$uv\neq 0$

On dividing each of the given equations by

$uv$ , we get

$\dfrac{8}{u}-\dfrac{3}{v}=5\quad\quad\quad\dots(i)$

$\dfrac{6}{u}-\dfrac{5}{v}=-2\quad\quad\quad\dots(ii)$

Taking

$\dfrac{1}{u}=x$ and

$\dfrac{1}{v}=y$ , the above equations become

$8x-3y=5\quad\quad\quad\dots(iii)$

$6x-5y=-2\quad\quad\quad\dots(iv)$

Multiplying equation

$(i)$ by

$3$ and equation

$(ii)$ by

$4$ , we get

$24x-9y=15\quad\quad\quad\dots(v)$

$24x-20y=-8\quad\quad\quad\dots(vi)$

Substituting equation

$(vi)$ from equation

$(v)$ , we get

$11y=23\Rightarrow y=\dfrac{23}{11}$

Putting

$y=\dfrac{23}{11}$ in equation (iii), we get

$8x-\dfrac{69}{11}=5$

$\Rightarrow 8x=\dfrac{69}{11}+5$

$\Rightarrow 8x=\dfrac{124}{11}$

$\Rightarrow x=\dfrac{31}{22}$

Now,

$x=\dfrac{31}{22}$

$\Rightarrow \dfrac{1}{u}=\dfrac{31}{22}$

$\Rightarrow u=\dfrac{22}{31}$

And,

$y=\dfrac{23}{11}$

$\Rightarrow \dfrac{1}{v}=\dfrac{23}{11}$

$\Rightarrow v=\dfrac{11}{23}$

Hence, the given system of equations has two solutions given by

$(i)$

$u=0, v=0$

$(ii)$

$u=\dfrac{22}{31}, v=\dfrac{11}{23}$ .

Solve:
$3(2u+v)=7uv$ and
$3(u+3v)=11uv$ .

Clearly, the given equations are not linear equations in the variables u and v but can be reduced to linear equations by appropriate substitution.

If we put

$u=0$ in either of the two equations, we get

$v=0$ .
So,

$u=0, v=0$ form a solution of the given system of equations.

To find the other solutions, we assume that

$u\neq 0, v\neq 0$ .
Now,

$u\neq 0, v\neq 0 \Rightarrow uv\neq 0$ .

On dividing each one of the given equations by uv, we get

$\dfrac{6}{v}+\dfrac{3}{u}=7$ (i)

$\dfrac{3}{v}+\dfrac{9}{u}=11$ (ii)

Taking

$\dfrac{1}{u}=x$ and

$\dfrac{1}{v}=y$ , the given equations become

$3x+6y=7$ ..(iii)

$9x+3y=11$ .(iv)

Multiplying equation (iv) by

$2$ , the given system of equations becomes

$3x+6y=7$ .(v)

$18x+6y=22$ .(vi)

Substracting equation (vi) from equation (v), we get

$-15x=-15\Rightarrow x=1$

Putting

$x=1$ in equation (iii), we get

$3+6y=7\Rightarrow y=\dfrac{4}{6}=\dfrac{2}{3}$

Now,

$x=1\Rightarrow \dfrac{1}{u}=1\Rightarrow u=1$

and,

$y=\dfrac{2}{3}\Rightarrow \dfrac{1}{v}=\dfrac{2}{3}\Rightarrow v=\dfrac{3}{2}$ .

Hence, the given system of equations has two solutions given by

(i)

$u=0, v=0$

(ii)

$u=1, v=3/2$ .

Solve the following system of equations for $x$ and $y$ ,

$\dfrac{a}{x}-\dfrac{b}{y}=0$ and

$\dfrac{ab^2}{x}+\dfrac{a^2b}{y}=a^2+b^2$ , where $x, y$ $\neq 0$ .

The given equations are,

$\dfrac{a}{x}-\dfrac{b}{y}=0$ and

$\dfrac{ab^2}{x}+\dfrac{a^2b}{y}=a^2+b^2$ , where

$x, y$

$\neq 0$ .

Taking

$\dfrac{1}{x}=u$ and

$\dfrac{1}{y}=v$ , the above system of equations becomes

$au-bv+0=0$

$ab^2u+a^2 bv-(a^2+b^2)=0$

By cross-multiplication, we have

$\dfrac{u}{(-b)\times [-(a^2+b^2)]-a^2b\times 0}=\dfrac{-v}{a\times [-(a^2+b^2)]-ab^2\times 0}=\dfrac{1}{a\times a^2b-ab^2\times (-b)}$

$\Rightarrow \dfrac{u}{b(a^2+b^2)}=\dfrac{-v}{-a(a^2+b^2)}=\dfrac{1}{a^3b+ab^3}$

$\Rightarrow \dfrac{u}{b(a^2+b^2)}=\dfrac{v}{a(a^2+b^2)}=\dfrac{1}{ab(a^2+b^2)}$

So,

$u=\dfrac{b(a^2+b^2)}{ab(a^2+b^2)}=\dfrac{1}{a}$

and,

$v=\dfrac{a(a^2+b^2)}{ab(a^2+b^2)}=\dfrac{1}{b}$

Now,

$u=\dfrac{1}{a}\Rightarrow \dfrac{1}{x}=\dfrac{1}{a}\Rightarrow x=a$

and

$v=\dfrac{1}{b}\Rightarrow \dfrac{1}{y}=\dfrac{1}{b}\Rightarrow y=b$

Hence, the solution of the given system of equations is

$x=a$ and

$y=b$ .

Solve the following system of equations by using the method of elimination by equating the coefficients:

$\dfrac{x}{10}+\dfrac{y}{5}+1=15$ and $\dfrac{x}{8}+\dfrac{y}{6}=15$ .

The given system of equations is

$\dfrac{x}{10}+\dfrac{y}{5}=14$

$\dfrac{x}{8}+\dfrac{y}{6}=15$

This system of equations can be re-written as

$x+2y=140$ ..(i)

$3x+4y=360$ .(ii)

Let us eliminate y from the equations (i) and (ii). The coefficients of y in the given equations are

$2$ and

$4$ respectively. The l.c.m of

$2$ and

$4$ is

$4$ .

$2x+4y=280$ ..(iii)

$3x+4y=360$ (iv)

Subtracting (iv) from (iii), we get

$-x=-80\Rightarrow x=80$

Putting

$x=80$ in equation (i), we get

$80+2y=140\Rightarrow 2y=60\Rightarrow y=30$

Hence, the solution of the given system of equations is

$x=80, y=30$ .

Solve the following systems of linear equations by using the method of elimination by equating the coefficients:
(i) $3x+2y=11$
(ii) $2x+3y=4$

The given systems of equations is

$3x+2y=11$

$(i)$

$2x+3y=4$

$(ii)$

Let us eliminate

$y$ from the given equations. The coefficients of

$y$ in the given equations are

$2$ and

$3$ respectively. The L.C.M. of

$2$ and

$3$ is

$6$ . So, we make the coefficients of

$y$ equal to

$6$ in the two equations.

Multiplying

$(i)$ by

$3$ and

$(ii)$ by

$2$ , we get

$9x+6y=33$ ...

$(iii)$

$4x+6y=8$ ..

$(iv)$

Subtracting

$(iv)$ from

$(iii)$ , we get

$5x=25\\\Rightarrow x=5$

Substituting

$x=5$ equation in

$(i)$ , we get

$15+2y=11\\\Rightarrow 2y=-4\\\Rightarrow y=-2$

Hence, the solution of the given system of equations is

$x=5, y=-2$ .

Solve the following systems of equations by using the method of substitution,
$2x+3y=9$ and $3x+4y=5$ .

The given system of equations is

$2x+3y=9$

$3x+4y=5$

From equation (i), we get

$3y=9-2x\Rightarrow y=\dfrac{9-2x}{3}$

Substituting

$y=\dfrac{9-2x}{3}$ in equation (ii), we get

$3x+4\left(\dfrac{9-2x}{3}\right)=5$

$\Rightarrow \dfrac{9x+36-8x}{3}=5$

$\Rightarrow x+36=15$

$\Rightarrow x=-21$

Putting

$x=-21$ in

$y=\dfrac{9-2x}{3}$ , we get

$y=\dfrac{9+42}{3}=17$

Hence, the solution of the given system of equations is

$x=-21, y=17$ .

Solve: $\dfrac{1}{2(2x+3y)}+\dfrac{12}{7(3x-2y)}=\dfrac{1}{2}$ and

$\dfrac{7}{2x+3y}+\dfrac{4}{3x-2y}=2$ , where $2x+3y\neq 0$ and $3x-2y\neq 0$ .

Let

$\dfrac{1}{2x+3y}=u$ and

$\dfrac{1}{3x-2y}=v$ . Then, the given system of equations becomes

$\dfrac{1}{2}u+\dfrac{12}{7}v=\dfrac{1}{2}\Rightarrow 7u+24v=7$ ..(i)

and,

$7u+4v=2$ .(ii)

Substracting equation (ii) from equation (i), we get

$20v=5\Rightarrow v=\dfrac{1}{4}$

Putting

$v=\dfrac{1}{4}$ in equation (i), we get

$7u+6=7\Rightarrow u=\dfrac{1}{7}$

Now,

$u=\dfrac{1}{7}\Rightarrow \dfrac{1}{2x+3y}=\dfrac{1}{7}\Rightarrow 2x+3y=7$ (iii)

and,

$u=\dfrac{1}{4}\Rightarrow \dfrac{1}{3x-2y}=\dfrac{1}{4}\Rightarrow 3x-2y=4$ .(iv)

Multiplying equation (iii) by

$2$ and equation (iv) by

$3$ , we get

$4x+6y=14$ .(v)

$9x-6y=12$ .(vi)

Adding equations (v) and (vi), we get

$13x=26\Rightarrow x=2$

Putting

$x=2$ in equation(v), we get

$8+6y=14\Rightarrow y=1$

Hence,

$x=2, y=1$ is the solution of the given system of equations.

For what value of k, will the following system of equations has infinitely many solutions?
$2x+3y=4$
$(k+2)x+6y=3k+2$

We know that the system of equations

$a_1x+b_1y=c_1$

$a_2x+b_2y=c_2$

has infinitely many solutions, if

$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$

Here,

$a_1=2, b_1=3, c_1=4, a_2=k+2, b_2=6,c_2=3k+2$ .

Therefore, the given system of equations will have infinitely many solutions, if

$\dfrac{2}{k+2}=\dfrac{3}{6}=\dfrac{4}{3k+2}$

$\Rightarrow \dfrac{2}{k+2}=\dfrac{3}{6}$ and

$\dfrac{3}{6}=\dfrac{4}{3k+2}$

$\Rightarrow \dfrac{2}{k+2}=\dfrac{1}{2}$ and

$\dfrac{1}{2}=\dfrac{4}{3k+2}$

$\Rightarrow k+2=4$ and

$3k+2=8$

$\Rightarrow k=2$ and

$k=2$

$\Rightarrow k=2$

Hence, the given system of equations will have infinitely many solutions, if

$k=2$ .

Solve $\dfrac{2}{x}+\dfrac{2}{3y}=\dfrac{1}{6}$ and $\dfrac{3}{x}+\dfrac{2}{y}=0$ and hence find 'a' for which, $y=ax-4$ .

As,

$\dfrac{2}{x}+\dfrac{2}{3y}=\dfrac{1}{6}$ and

$\dfrac{3}{x}+\dfrac{2}{y}=0$ are not in the general form of linear equations. To make them into linear equaions,

Take

$\dfrac{1}{x}=u$ and

$\dfrac{1}{y}=v$ . The given system of equations become

$2u+\dfrac{2}{3}v=\dfrac{1}{6}$

$\Rightarrow 12u+4v=1$ (i)
and,

$3u+2v=0$ .(ii)

Multiplying equation (ii) by

$2$ and subtracting from equation (i), we get

$6u-1\Rightarrow u=\dfrac{1}{6}$

Putting

$u=\dfrac{1}{6}$ in (i), we get

$2+4v=1\Rightarrow v=-\dfrac{1}{4}$

Hence,

$x=\dfrac{1}{u}=6$ and

$y=\dfrac{1}{v}=-4$

So, the solution of the given system of equations is

$x=6, y=-4$

Putting

$x=6, y=-4$ in

$y=ax-4$ , we get

$-4=6a-4\Rightarrow a=0$ .

Solve the following system of equations,

$\dfrac{1}{2x}-\dfrac{1}{y}=-1$ and $\dfrac{1}{x}+\dfrac{1}{2y}=8$ , where $x\neq 0, y\neq 0$ .

Taking

$\dfrac{1}{x}=u$ and

$\dfrac{1}{y}=v$ , the given equations become

$\dfrac{u}{2}-v=-1\Rightarrow u-2v=-2$ .(i)

and,

$u+\dfrac{v}{2}=8\Rightarrow 2u+v=16$ .(ii)

Let us eliminate u from equations (i) and (ii). Multiplying equation (i) by

$2$ , we get

$2u-4v=-4$ ..(iii)

$2u+v=16$ (iv)

Subtracting (iv) from (iii), we get

$-5v=-20\Rightarrow v=4$

Putting

$v=4$ in equation (i), we get

$u-8=-2\Rightarrow u=6$

Hence,

$x=\dfrac{1}{u}=\dfrac{1}{6}$ and

$y=\dfrac{1}{v}=\dfrac{1}{4}$
So, the solution of the given system of equation is

$x=\dfrac{1}{6}, y=\dfrac{1}{4}$ .

Find solution of a pair of linear equations in two variables by elimination method
$2x+3y=0$
$3x+4y=5$

Given:

$2x+3y=0\longrightarrow (i)\\ 3x+4y=5\longrightarrow (ii)\\$

Multiplying

$(i)$ by

$3$ and

$(ii)$ by

$2$

$6x+9y=0\\ 6x+8y=10$

On subtraction the equations:

$y=-10$

putting the value of

$y$ in

$(i)$

$\Rightarrow 2x+3(-10)=0$

$\Rightarrow x=15$

Solve for $x$ and $y$
$2x-3y=8$
$5x$ $-7y=19$

$2x-3y=8\longrightarrow (i)\\ \therefore \quad x=\cfrac { 8+3y }{ 2 } \\$

Putting value of

$x$ in equation:

$5x-7y=19\longrightarrow (ii)\\ \Rightarrow 5\left( \cfrac { 8+3y }{ 2 } \right) -7y=19\\ \Rightarrow 5\left( 8+3y \right) -14y=38\\ \Rightarrow y=-2$

Putting value of

$y$ in equation

$\longrightarrow(i)$

$2x+6=8$

$x=1$

Solve: $\dfrac{20}{x+y}+\dfrac{3}{x-y}=7$ , $\dfrac{8}{x-y}-\dfrac{15}{x+y}=5$

Let

$\dfrac{1}{x+y}=u \dfrac{1}{x-y}=v$

$204+3v=7$ and

$8v-15u=5$

$24v+160\mu =56$

$24v-45\mu =15$

$-$

$+$

$-$

$\overline{\quad \quad \quad \, 205 \mu = 41}$

$u=\dfrac{41}{205}=\dfrac{1}{5}$

$v=1$

$\dfrac{1}{x+y}=\dfrac{1}{5}$

$\dfrac{1}{x-y}=1$

$x+y=5$

$x-y=1$

$\overline{\quad 2x=6}$

$x=3$

$y=2$

Solve for $'x'$ and $'y'$ $2x-3y=13,7x-2y=20$

$2x-3y=13\longrightarrow (i)\\ 7x-2y=20\longrightarrow (ii)\\$

Multiplying

$(i)$ by

$2$ and

$(ii)$ by

$3$

$4x-6y=26\\ 21x-6y=60$

On subtraction :

$-17x=-34$

$x=2$

putting the value of

$x$ in

$(i)$

$2\left( 2 \right) -3y=13\\ \Rightarrow -3y=9\\ \therefore \quad y=-3$

Solve:
$5x-4y+8=0$
$7x+6y-9=0$

$(5x−4y+8=0)\times3$

$15x-12y+24=0............(1)$

$(7x+6y−9=0)\times2$

$14x+12y-18=0............(2)$

Addition of equation

$(1)$ and

$(2)$ we get,

$29x+6=0$

$\Rightarrow x=\dfrac{-6}{29}$

Substitute value of

$x$ in equation

$5x-4y+8=0,$

$5\left(\dfrac{-6}{29}\right)-4y+8=0$

$\Rightarrow 4y=8- \dfrac{30}{29}$

$\Rightarrow 4y=\dfrac{202}{29}$

$\Rightarrow y=\dfrac{101}{58}$

Hence

$x=\dfrac{-6}{29},\ y=\dfrac{101}{58}$ is the solution of given pair of equations.

Solve graphically the system of linear equation: $x+3y=6$ , $2x-3y=12$

$x+3y=6\\$

X	6	3	0
Y	0	1	2

$2x-3y=12$

X	0	6	3
Y	-4	0	-2

1016589_1036102_ans_ed4d1a7402ce48ae80595e4e2a6d9ed5.png

Solve the following systems of linear equations by using the method of elimination by equating the coefficients:
$8x+5y=9$
$3x+2y=4$

The given system of equation is

$8x+5y=9$ .....(i)

$3x+2y=4$ .....(ii)

Let us eliminate

$x$ from the given equations. The coefficients of

$x$ in the given equations are

$8$ and

$3$ respectively. The L.C.M. of

$8$ and

$3$ is

$24$ . So, we make both the coefficients equal to

$24$ .

Multiplying both sides of equation (i) by

$3$ and equation (ii) by

$8$ , we get:

$24x+15y=27$

$24x+16y=32$

Substracting (iv) from (iii), we get:

$−y=−5 \implies y=5$

Putting

$y=5$ in (i), we get:

$8x+25=9 \implies 8x=−16$

$\implies x=−2$

Hence, the solution of the given system of equations is

$x=−2,\ y=5$ .

If the pair of integers whose sum is zero but difference is $12$ are $x\ and\ y$ . Then $xy=-m$ .Find $m$

$x + y = 0 \longrightarrow \left( i \right)$

$x - y = 12 \longrightarrow \left( ii \right)$

Adding

${eq}^{n} \left( i \right) \& \left( ii \right)$ , we have

$x + y + x - y = 0 + 12$

$\Rightarrow \; 2x = 12$

$\Rightarrow \; x = 6$

Substituting the value of x in

${eq}^{n} \left( i \right)$ , we have

$6 + y = 0$

$\Rightarrow \; y = -6$

$\therefore \; xy = 6 \times \left( -6 \right) = -36 \longrightarrow \left( iii \right)$

$xy = -m \longrightarrow \left( iv \right) \; \left\{ Given \right\}$

From

${eq}^{n} \left( i \right) \& \left( ii \right)$ , we get

$m = 36$

Solve the following systems of equations by using the method of substitution:
$x+2y=-1$
$2x-3y=12$

The given system of equations is

$x+2y=-1$ (i)

$2x-3y=12$ .(ii)

From equation (i), we get

$x=-1-2y$

Substituting

$x=-1-2y$ in equation (ii), we get

$\Rightarrow 2(-1-2y)-3y=12$

$\Rightarrow -2-4y-3y=12$

$\Rightarrow -7y=14$

$\Rightarrow y=-2$

Putting

$y=-2$ in

$x=-1-2y$ , we get

$x=-1-2\times (-2)=3$

Hence, the solution of the given system of equations is

$x=3, y=-2$ .

Find out whether the lines representing the following pairs of linear equation intersect at a point, are parallel or coincident: $5x-4y+8=0$ and $7x+6y-9=0$

$\cfrac{{{a_1}}}{{{a_2}}} = \cfrac{5}{7}$

$\cfrac{{{b_1}}}{{{b_2}}} = \cfrac{{ - 4}}{6} = \cfrac{{ - 2}}{3}$

Since

$\cfrac{{{a_1}}}{{{a_2}}} \ne \cfrac{{{b_1}}}{{{b_2}}}$

Therefore given pair of lines intersect at a point

It can take $12$ hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for four hours and the pipe of smaller diameter for $9$ hours, only half of the pool can be filled. How long would it take for each pipe to fill the pool separately?

Let there are two pipes A and B and diameter of A is larger than B.

Now suppose that

Pipe A take

$x$ hours and Pipe B takes

$y$ hours to fill the pool separately .

In 1 hour pipe A can fill the pool

$=\dfrac{1}{x}$

In 1 hour pipe B can fill the pool

$=\dfrac{1}{y}$

If both pipe are together then they take

$12$ hour to fill the pool.

If the are togther then in 1 hour they fill the pool.

$\dfrac{1}{x}$ +

$\dfrac{1}{y}$ =

$\dfrac{1}{12}----(1)$

Now

$4$ hour pipe A can fill the pool is

$=\dfrac{4}{x}$

$9$ hour pipe B can fill the pool is

$=\dfrac{9}{y}$

than they fill half of the pool,so

$\dfrac{4}{x}$ +

$\dfrac{9}{y}$

$=\dfrac{1}{2}-----(2)$

Now Let

$\dfrac{1}{x}=P$ &

$\dfrac{1}{y}=Q$ and puuting in equation 1 &2, than

$P+Q$ =

$\dfrac{1}{12}$

$\Rightarrow12P+12Q=1----(3)$

And

$4P+9Q$

$=\dfrac{1}{2}$

$\Rightarrow8P+18Q=1-----(4)$

Now solving equation 3 & 4

Equation (3) multiplying by 2 and (4) multiplying by 3 and substract.

$\Rightarrow -30Q=-1$

$\Rightarrow Q$

$=\dfrac{1}{30}$

now put value of

$Q$ in equation

$3$

We get

$P$

$=\dfrac{1}{20}$

So we found

$P$

$=\dfrac{1}{20}$ &

$Q$

$=\dfrac{1}{30}$

$x=20$ and

$y=30$

Hence the A pipe would take 20 hours and the pipe B would take 30 hours separately .

Solve: $3x+4y=25, \ 5x-6y=-9$ by elimination method.

Given:

$3x + 4y = 25$ ....... (1)

$5x - 6y = - 9$ ....... (2)

Multiply equation (1) by 3 and equation (2) by 2 we get,

$9x + 12y = 75$

$10x - 12y = - 18$

Now, by adding both the equation we get,

$19x = 57$

$x = 3$

Substitute $x = 3$ in equation (1) we get,

$3\times 3+ 4y = 25$

$\Rightarrow y=\dfrac{16}{ 4} = 4$

Solve: $3x+2y=15$ and $10x-4y=2$

Given equations can be written as,

$3x + 2y -15= 0\quad\quad\quad\dots(i)$

$10x - 4y-2 = 0\quad\quad\quad\dots(ii)$

Multiply

$(i)$ by

$2$ and add it in equation

$(ii),$

$\begin{aligned}{}2\left( {3x + 2y - 15} \right) + \left( {10x - 4y - 2} \right)& = 0\\6x + 4y - 30 + 10x - 4y - 2 &= 0\\16x - 32& = 0\\16x& = 32\\x &= 2\end{aligned}$

Substitute

$x=2$ in equation

$(i),$

$\begin{aligned}{}3 \times 2 + 2y& = 15\\6 + 2y &= 15\\2y &= 15 - 6\\2y& = 9\\y& = \frac{9}{2}\end{aligned}$

Therefore

$x=2$ and

$y = \dfrac{9}{2}$ is the solution for the given pair of equation.

Find the values of $'a'$ and $'b'$ for which the given pair of linear equations $2x+3y=7;2ax+(a+b)=28$ have infinite number of solutions.

$2x + 3y = 7$

$2ax + \left( {a + b} \right)y = 28$

For infinitely many solutions

$\cfrac{2}{{2a}} = \cfrac{3}{{a + b}} = \cfrac{7}{{28}}$

$\Rightarrow \cfrac{1}{a} = \cfrac{3}{{a + b}} = \cfrac{1}{4}$

$\Rightarrow \cfrac{1}{a} = \cfrac{1}{4}$ and

$\cfrac{3}{{a + b}} = \cfrac{1}{4}$

$\Rightarrow a = 4$ and

$a + b = 12$

Now

$a + b = 12$

$\Rightarrow 4 + b = 12$

$\Rightarrow b = 8$

Therefore for infinite number of solutions

$a=4$ and

$b=8$

For which value of $k$ will the following pair of linear equations have no solution?
$3x+y=1$ and $(2k-1)x+(k-1)y=2k+1$

We know that if

$\dfrac{{a}_{1}}{{a}_{2}}=\dfrac{{b}_{1}}{{b}_{2}}\neq\dfrac{{c}_{1}}{{c}_{2}}$ then the sysytem of linear equations has no solution.

where

${a}_{1}=3,\,{b}_{1}=1,\,{a}_{2}=2k-1$ and

${b}_{2}=k-1$

$\Rightarrow\,\dfrac{3}{2k-1}=\dfrac{1}{k-1}$

$\Rightarrow\,\dfrac{3}{2k-1}-\dfrac{1}{k-1}=0$

$\Rightarrow\,\dfrac{3k-3-2k+1}{\left(2k-1\right)\left(k-1\right)}=0$ for

$k\neq 1,\dfrac{1}{2}$

$\Rightarrow\,k-2=0$

$\therefore\,k=2$

Solve graphically : $y = 2x + 1$ and $x = 2y - 5$

$x=2y-5$

x	-3	-1	1
y	1	2	3

$y=2x+1\\$

x	0	1	-1
y	1	3	-1

The graph has been plotted using the above points and the point of intersectio comes out to be

$(1,3)$ .

1016915_1049959_ans_a9023652442f4954ba25533d965cfe18.png

Solve:
$4x+\dfrac {x-y}{8}=17$

$2y+x-\dfrac {5y+2}{3}=2$

$4x + \dfrac{{x - y}}{8} = 17$

$\implies 32x+x-y=17\times 8$

$\implies 33x-y=136$ ------- (1)

$2y + x - \dfrac{{5y + 2}}{3} = 2$

$\implies 3\times (2y+x)-(5y+2)=6$

$\implies 3x+y=4$ ------ (2)

$\implies y=4-3x$ ------ (3)

put in (1)

$\implies 33x-4+3x=136$

$\implies 36x=140$

$\implies x=\dfrac{140}{36}$

from (3)

$y=4-3\times \dfrac{140}{36}$

$\implies y=4-\dfrac{140}{12}$

$\implies y=\dfrac{-92}{12}$

Solve the following pairs of equations by reducing them to a pair of linear equations:
$\dfrac{1}{2x}+\dfrac{1}{3y}=2$ , $\dfrac{1}{3x}+\dfrac{1}{2y}=\dfrac{13}{6}$

Given equations are

$\dfrac{1}{2x}+\dfrac{1}{3y}=2$ and

$\dfrac{1}{3x}+\dfrac{1}{2y}=\dfrac{13}{6}$

Put

$\dfrac{1}{x}=u,\dfrac{1}{y}=v$

$\Rightarrow \dfrac{1}{2}u+\dfrac{1}{3}v=2\:and\: \dfrac{1}{3}u+\dfrac{1}{2}v=\dfrac{13}{6}$

$\Rightarrow 3u+2v=12 \:and\: 2u+3v=13$

Solving the above equations we get

$u=2,v=3$

Hence

$\dfrac{1}{x}=u=2 \Rightarrow x=\dfrac{1}{2}$ and

$\dfrac{1}{y}=v=3 \Rightarrow y=\dfrac{1}{3}$

Therefore

$x=\dfrac{1}{2},y=\dfrac{1}{3}$

Write the condition for pair of linear Equation having Infinite number or solutions
$a_{1}x + b_{1}y + c_{1} = 0$ and $a_{2}x + b_{2}y + c_{2} = 0$ .

$a_{1}x + b_{1}y + c_{1} = 0$ and

$a_{2}x + b_{2}y + c_{2} = 0$

will have infinite Number of Solutions when

$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$

By elimination method find the solution set of the given pair $4x-3y=5$ and $7x-15y=-1$

Let

$4x-3y=5$ ------ (1)

$7x-15y=-1$ ----- (2)

Multiply (1) by

$5$

We get,

$20x-15y=25$ ------ (3)

Let , (2) - (3)

We get,

$x=2$ put in (1)

$y=1$

Solve graphically: $2x-y-2=0$ , $4x-y-4=0$

Graph of the equation of given two lines intersect at

$(1,0)$ hence, it is the solution of the given equations.
1160019_1058969_ans_15647bb0dd334f8482b8415350b785af.png

What is meant by consistent? What is meant by inconsistent?

If atleast one set of values occurred for the unknowns that satisfies every equation in the system, then that system of equations is known as consistent.

If no set of values satisfies the equation, then that system is known as inconsistent.

Solve the pair of linear equations $2x+3y=11$ and $2x-4y=-24$ .

The pair of linear equations is given as,

$2x + 3y = 11\quad\quad\quad\dots(i)$

$2x - 4y = - 24\quad\quad\quad\dots(ii)$

Subtracting equation $(ii)$ from equation $(i),$

$(2x+3y)-(2x-4y)=11-(-24)$

$\Rightarrow 7y = 35$

$\Rightarrow y = 5$

Substitute the value of $y$ in equation $(i),$

$2x + 3\left( 5 \right) = 11$

$\Rightarrow 2x + 15 = 11$

$\Rightarrow 2x = - 4$

$\Rightarrow x = - 2$

Therefore, the values are $x = - 2$ and $y = 5$ .

Solve
$3x+2y=8$
$2x-3y=1$

$3x + 2y = 8 - \left( 1 \right)$

$2x - 3y = 1 - \left( 2 \right)$

Multiply equation

$(1)$ by

$3$ and

$(2)$ by

$2$ we get,

$9x + 6y = 24 - \left( 3 \right)$

$4x - 6y = 2 - \left( 4 \right)$

Adding equations

$(2)$ and

$(4)$ we get

$\left( {9x + 6y} \right) + \left( {4x - 6y} \right) = 24 + 2$

$\Rightarrow 13x = 26$

$\Rightarrow x = 2$

Substitute

$x=2$ in equation

$(1),$

$3x + 2y = 8$

$\Rightarrow 3 \times 2 + 2y = 8$

$\Rightarrow 6 + 2y = 8$

$\Rightarrow 2y = 2$

$\Rightarrow y = 1$

Hence, the solution for the given system of equation is

$(2,1).$

Solve for $x$ and $y$ if $3(2x+y)=7xy$ and $3(x+3y)=11x$ .

$3\left( {2x + y} \right) = 7xy$

$\Rightarrow$

$6x + 3y = 7xy$

$\Rightarrow$

$\dfrac{6}{y} + \dfrac{3}{x} = 7$

$\Rightarrow$

$\dfrac{3}{x} + \dfrac{6}{y} = 7$ ......

$\left( 1 \right)$

$\Rightarrow$

$3\left( {x + 3y} \right) = 11x$

$\Rightarrow$

$3x + 9y = 11$

$\Rightarrow$

$8x = 9y$

$\Rightarrow$

$x = \dfrac{{9y}}{8}$

Putting

$x = \dfrac{{9y}}{8}$ in

$(1)$ , we get

$\dfrac{{3 \times 8}}{{9y}} + \dfrac{6}{y} = 7$

$\dfrac{1}{y}\left( {\dfrac{8}{3} + 6} \right) = 7$

$7y = \dfrac{{26}}{3}$

$y = \dfrac{{26}}{21}$

From

$(1)$ , we get

$x = \dfrac{{9y}}{8} = \dfrac{9}{8} \times \dfrac{{26}}{{21}}$

So,

$x=\dfrac{{39}}{{28}}$ and

$y= \dfrac{{26}}{{21}}$

Solve the equation by elimination method
$11x + 15y + 23 = 0$ , $7x - 2y - 20 = 0$

$2\times 11x+15y+23=0 \implies 22x+30y+46=0$

$15\times 7x-2y-20=0 \implies 105x-30y-300=0$

adding both

$127x-254=0 \implies x=2$

$y=-3$

Find the value of x and y:
$5x-3y=11$ , $10x+6y=-22$

Let the equation

$5x - 3y = 11$ ----- (i)

$10x+6y=-22$ ----- (ii)

From (i)

$y=\dfrac{5x-11}{3}$

put the value of

$y$ in (ii)

$10x+6(\dfrac{5x-11}{3})=-22$

$30x+30x-66=-66$

$x=0$

put

$x=0$ in (i), we get

$0-3y=11$

$y=-\dfrac{11}{3}$

If $-2x=4y+6$ and $2(2y+3)=3x-5$ , what is the solution $(x,y)$ to the system of equations above?

The given set of equations are,

$- 2x = 4y + 6.....(i)$

and

$2\left( {2y + 3} \right) = 3x - 5.......(ii)$

from (i)

$- 2x = 4y + 6$

$\Rightarrow 2x + 4y = - 6......(iii)$

And,from (ii)

$2\left( {2y + 3} \right) = 3x - 5$

$\Rightarrow 4y + 6 = 3x - 5$

$\Rightarrow 3x - 4y = 11.......(iv)$

adding (iii) and (iv)

$5x=5$

$\Rightarrow x=1$

from (iii)

$2\times 1+4y=-6$

$\Rightarrow2+4y=-6$

$\Rightarrow 4y=-8$

$\Rightarrow y=-2$

Therefore, the solution to the given system of the equations is

$\left( {1, - 2} \right)$ .

Draw the graph for each of the equation $x + y = 6$ and $x - y = 2$ on the same graph paper and find the coordinates of the point where the two straight lines intersect.

As we can see from the graph that the two given lines that there are

$x+y=6$ and

$x-y=2$ intersect at

$(4,2)$

1424542_1066789_ans_1e3be473b4244368b5fd369b3f0ea9e3.png

Solve :
$\dfrac{5}{x + y} - \dfrac{2}{x - y} + 1 = 0 , \\ \, \dfrac{15}{x + y} + \dfrac{7}{x - y } -10 = 0$

$\cfrac{5}{x+y}-\cfrac{2}{x-y}=-1$ .................. (1)

$\cfrac{15}{x+y}+\cfrac{7}{x-y}=10$ ................. (2)

Let

$\cfrac{1}{x+y}=u$ and

$\cfrac{1}{x-y}=v$

Then equation (1) and (2) becomes,

$5u-2v=-1$ ................. (3)

$15u+7v=10$ ...............(4)

Multiply equation (3) with

$3$ ,

$15u-6v=-3$ ................. (5)

Now subtract the above result from equation (4),

$7v+6v=10+3$

$13v=13$

$v=1$

$\Rightarrow x-y=1$ ................... (6)

From equation (3),

$5u=-1+2v$

$5u=-1+2$

$5u=1$

$u=\cfrac{1}{5}$

$\Rightarrow x+y=5$ ................... (7)

Adding (6) and (7) , we get,

$2x=6$

$x=3$

From equation (7),

$y=5-x$

$y=5-3$

$y=2$

Solve the following pairs of equations by reducing them to a pair of linear equations:
$6x+3y=6xy$ $2x+4y=5xy$

The given pair of equations are

$6x+3y=6xy,2x+4y=5xy$

Divide first equation by

$3$ we get

$2x+y=2xy\quad ...(1)$ and

$2x+4y=5xy\quad ...(2)$

Divide

$(1),(2)$ by

$xy$ we get

$\dfrac{2}{y}+\dfrac{1}{x}=2\quad ...(3)$ and

$\dfrac{2}{y}+\dfrac{4}{x}=5\quad ...(4)$

Let

$\dfrac{1}{x}=p,\dfrac{1}{y}=q$ we get

$p+2q=2\quad ...(5)$ and

$4p+2q=5\quad ...(6)$

$(5) \Rightarrow p=2-2q$

Substitute

$p=2-2q$ in

$(6)$ we get

$4(2-2q)+2q=5$

$\Rightarrow 8-8q+2q=5$

$\Rightarrow 8-6q=5$

$\Rightarrow 6q=3$

$\Rightarrow q=\dfrac{1}{2}$

Hence

$p=2-2q=2-2\times \dfrac{1}{2}=2-1=1$

Therefore

$\dfrac{1}{x}=p=1 \Rightarrow x=1$ and

$\dfrac{1}{y}=q=\dfrac{1}{2} \Rightarrow y=2$

Solution:

$x=1,y=2$

Solve graphically: $x+3y=6$ and $2x+3y=12$ .

REF.Image.

Let straight

A : x+3y = 6 &

B = 2x+3y = 12

For A

x	0	6
y	2	0

For B

x	0	6
y	4	0

From the graph we can see that at point (6,0), both lines are intersecting.

1216199_1068963_ans_7f26adce091c44dbaf4e658e4755aa05.jpg

We have a linear equation $2x+3y-8=0$ . Write another linear equation in two variables x and y such that the geometrical representation of the pair so formed is intersecting lines.

Equation of first line is

$2x+3y-8=0$

Let equation of second line be

$ax+by+c=0$

For intersecting lines:

$\dfrac{{a}_{1}}{{a}_{2}}\neq\dfrac{{b}_{1}}{{b}_{2}}$

Here

${a}_{1}=2,\,\,{b}_{1}=3,\,\,{a}_{2}=a,\,\,{b}_{2}=b$

Hence

$\dfrac{2}{a}\neq \dfrac{3}{b}$ .......

$(1)$

So, we assume a value of

$a$ and

$b$ such that equation

$(1)$ is satisfied.

Let equation of second line be

$3x+2y+4=0$

Here

$a=3,\,b=2,\,c=4$ and equation is satisfied

$\dfrac{2}{3}\neq \dfrac{3}{2}$

Solve the following pairs of equations by reducing them to a pair of linear equations:

$\dfrac{2}{\sqrt{x}}+\dfrac{2}{\sqrt{y}}=2$ , $\dfrac{4}{\sqrt{x}}-\dfrac{9}{\sqrt{y}}=-1$

Given equations are

$\dfrac{2}{\sqrt{x}}+\dfrac{2}{\sqrt{y}}=2$ and

$\dfrac{4}{\sqrt{x}}-\dfrac{9}{\sqrt{y}}=-1$

Put

$\dfrac{1}{\sqrt{x}}=u,\dfrac{1}{\sqrt{y}}=v$

$\Rightarrow 2u+2v=2\:and\: 4u-9v=-1$

Solving the above equations we get

$u=\dfrac{8}{13},v=\dfrac{5}{13}$

Hence

$\dfrac{1}{\sqrt{x}}=u=\dfrac{8}{13} \Rightarrow \sqrt{x}=\dfrac{13}{8} \Rightarrow x=\dfrac{169}{64}$ and

$\dfrac{1}{\sqrt{y}}=v=\dfrac{5}{13} \Rightarrow \sqrt{y}=\dfrac{13}{5} \Rightarrow y=\dfrac{169}{25}$

Therefore

$x=\dfrac{169}{64},y=\dfrac{169}{25}$

$\frac{5}{x+y}-\frac{2}{x-y}=-1$
$\frac{15}{x+y}+\frac{7}{x-y}=10$

$\\let\>(\frac{1}{x+y})=u\>and\>(\frac{1}{x-y})=v\\\therefore\>5-2v=-1\rightarrow\>(1.)\\15+7v=10\rightarrow\>(2.)\\multiply\>equation\>(1.)by\>.3,we\>get\>15u-6v=-3\rightarrow\>(1.)\\now\>subtract\>eq^n\>(2.)in\>to\>eq^n(1.)we\>get\>-6v-7v=-3-10\\=-13v\>=-13\\\therefore\>v=1\>or\>x+y=1\rightarrow\>(3.)\\and\>u=(\frac{-1+2v}{5})\\=(\frac{-1+2}{5})=(\frac{+1}{5})\therefore\>x-y\>=5\rightarrow\>x=3(4.)Add\>equation\>(3.)and\>(4.),we\>get\>\\2x=1+5\>\therefore\>\\\>then=x-5=3-5=-2$

Solve the following pairs of equations by reducing them to a pair of linear equations:
$\dfrac{4}{x}+3y=14$ , $\dfrac{3}{x}-4y=23$

The given pair of equations are

$\dfrac{4}{x}+3y=14$ and

$\dfrac{3}{x}-4y=23$

Put

$\dfrac{1}{x}=u$ we get

$4u+3y=14 \quad ...(1),3u-4y=23 \quad ...(2)$

From

$(1)$ we get

$u=\dfrac{14-3y}{4}$

Substitute

$u$ in

$(2)$ we get

$3\left(\dfrac{14-3y}{4}\right)-4y=23$

$\Rightarrow 42-9y-16y=92$

$\Rightarrow -25y=50$

$\Rightarrow y=-2$

Substitute

$y=-2$ in

$u=\dfrac{14-3y}{4}$ we get

$u=5$

Hence

$\dfrac{1}{x}=u=5 \Rightarrow x=\dfrac{1}{5}$

Solution:

$x=\dfrac{1}{5},y=-2$

Draw the graph of i) $y = 2x + 5$ , ii) $y = 2x - 5$ , iii) $y = 2x$ and find the point of intersection on $x$ -axis. Is the $x$ -coordinates of these points also the zero of the polynomial ?

$y=2x+5$

$x=0,y=5$

$x=2,y=9$

$y=5,x=0$

$y=0,x=2.5$

ii)

$y=2x-5$

$x=0,y=-5$

$x=2,y=-1$

$x=3,y=1$

iii)

$y=2x$

$x=0,y=0$

$x=1,y=2$

$x=2,y=4$

$x=3,y=6$

1097233_1103324_ans_9c067831a1764ec8aa7d2ceecb507956.png

Solve the following pair of linear equation by the substitution method.
$0.2x+0.3y=1.3$
$0.4x+0.5y=2.3$

Given

$0.2x+0.3y=1.3---(1)$

$0.4x+0.5y=2.3---(2)$

from equation

$(1)$

$x=\dfrac {1.3-0.3y}{0.2}\quad$

substitute

$x$ in equation

$(2)$

$0.4 \left (\dfrac {1.3-0.3y}{0.2}\right) +0.5y=2.3$

$\dfrac {0.5\times 2-0.12y}{0.2}+0.5y=2.3$

$\dfrac {0.5\times 2-0.12y+0.10y}{0.2}=2.3$

$0.52-0.12y+0.10y=0.46$

$-0.02y=0.46-0.52$

$-0.02y=0.06$

$y=\dfrac {0.06}{0.02}$

$y=3$

Substitute

$y=3$ in

$x$

$x=\dfrac {1.8-0.3 (3)}{0.2}$

$=\dfrac {1.3 -0.9}{0.2}$

$=\dfrac {0.4}{0.2}$

$=2$

$\therefore \ x=2, y=3$

I have a total of $Rs.\ 300$ in coins of denomination $Rs.\ 1$ , $Rs.\ 2$ , $Rs.\ 5$ . The number of $Rs.\ 2$ coins is $3$ times the number of $Rs.\ 5$ coins. The total number of coins is $160$ . How many coins of each denomination are with me?

Let

$O,T,F$ be the number of

$Rs.\ 1,\ Rs.\ 2,\ Rs.\ 5$ coins.

Given that,

$T=3F$ .......

$(1)$

$O+T+F=160$ .........

$(2)$

Substitute

$(1)$ in

$(2)$ we get,

$O+3F+F=160$

$\Rightarrow O+4F=160$

$\Rightarrow O=160-4F$ .....

$(3)$

Also,

$\Rightarrow O+2T+5F=300$

Substitute the values from

$(1)$ and

$(3)$ in the above equation,

$160-4F+2(3F)+5F=300$

$\Rightarrow 7F=140$

$\Rightarrow F=20$

Thus,

$T=3F$

$\Rightarrow T=3\times 20$

$\Rightarrow T=60$

And,

$O=160-4(20)$

$\Rightarrow O=80$

So, there are

$80$ coins of

$Rs.\ 1,$

$60$ coins of

$Rs.\ 2$ and

$20$ coins of

$Rs.\ 5.$

Graphical seclusion: Solve : $x+y=10$
$x+y= 8$

$x+y=10\\\therefore\>x=0,\>y\>=10\>hence\>point\>(0,10)\\and\>if\>x\>=10,\>y\>=0\>hence\>point\>(10,0)\\for\>x+y=8\\at\>x\>=0,\>y=8,\>so\>point\>(0,8)\\and\>x=8,\>y\>=0,\>so\>point\>(8,0)$

As both line are parallel to each other and hence there are no solution to the given system of equations.

1184743_1073002_ans_0fb24774321a42dfa5dc890813a4e7d2.PNG

The present age of sahil's mother is three times the present age of sahil. After $5$ years their ages add to $66$ years find their present age.

Let present age of Sahil

$=x$ years

Let present age of Sahil's mother

$=y$ years

The present age of Sahil's mother is three times the present age of Sahil.

So,

$y=3x$ ......... (1)

After

$5$ years their ages add to

$66$ years.

After

$5$ years Sahil age

$=x+5$

After

$5$ years Sahil's mother age

$=y+5$

Now,

$x+5+y+5=66$

$x+y+10=66$

$x+y=66-10$

$x+y=56$ ......... (2)

From (1) and (2),

$x+3x=56$

$4x=56$

$x=14$ years

From (1),

$y=3x$

$=3(14)$

$=42$ years
present age of Sahil

$=14$ years

present age of Sahil's mother

$=42$ years

Solve for x and y:
$\frac{3x}{2}-\frac{5y}{3}+2=0$
$\frac{x}{3}+\frac{y}{2}=2\frac{1}{6}$

$(\frac{3x}{2})-(\frac{5y}{3})+2=0\rightarrow\>(1.)\\(\frac{x}{3})+(\frac{y}{2})=(\frac{13}{16})\rightarrow\>(2.)\\multiply\>equation\>(1)by\>(\frac{2}{9}),we\>get\>\\(\frac{x}{3})-(\frac{10y}{27})+(\frac{4}{9})=0\rightarrow\>(1.)\\now\>subtract\>eq^n(1)-(2),\>we\>get\>\\(\frac{-10y}{27})-(\frac{y}{2})+(\frac{4}{9})-(\frac{13}{6})=0\\(\frac{8-39}{18})=(\frac{20y+27y}{54})\\(\frac{-31}{18})=(\frac{47y}{54})\\\therefore\>y=(\frac{-31\times\>3}{47})=(\frac{-93}{47})\\\therefore\>x=((\frac{13}{6})-(\frac{y}{2}))\times\>3=((\frac{13}{6})+(\frac{93}{94}))\times\>\\=(\frac{(1222+558)\times\>3}{6\times\>94})=(\frac{890}{94})=(\frac{445}{47})$

Solve following equation by elimination method
$3x+2y=11$
$2x+3y=4$

Given that:

$3x+2y=11,$ ......(1)

On multiplying by

$3$ , we get

$9x+6y=33$ ......(2)

$2x+3y=4,$ ......(3)

On multiplying by

$2$ , we get

$4x+6y=8$ ......(4)

On subtracting equation (4) from equation (2), we get

$5x=25,$

$x=5$

Substituting in equation (1), we get

$15+2y=11$

$y=-2$

There are some $20$ paise and some $25$ paise coins in a bag. The total number of coins is $50$ and the amount in the bag is $Rs.\ 11.25$ . Find the number of coins of each value in the bag by method of elimination?

Let no. of 20 paise coins is x and no. of 25 paise coins is y. Then by the given question,

$x+y=50$ (equation 1)

$20x+25y=1125$

or, $4x+5y=225$ (equation 2)

Multiplying (1) with 4 we get,
$4x+4y=200$ (equation 3)

Subtracting (3) from (2) we get,
$y=25$

Putting in (1)

$x=50-y=50-25=25$

∴ $x=25, y=25$

Solve the system of equations
$2x + 5y = 1$
$3x + 2y = 7$ .

Consider the given equations.

$2x + 5y = 1 \quad \quad .... (1)$

$3x + 2y = 7 \quad \quad .... (2)$

On multiply by

$(1)$ by

$2$ and

$(2)$ by

$5$ , we get:

$4x + 10y = 2 \quad \quad \ \ \ \ .... (3)$

$\underline {\underset {-}{15}x \underset {-}{+} 10y \underset {-}{=} 35} \quad \quad .... (4)$

$-11x = - 33$ [Subtracting

$(4)$ from

$(3)$ ]

$\Rightarrow x = 3$

Putting

$x = 3$ in

$(1)$ , we get:

$2\times 3 + 5y = 1$

$\Rightarrow 6 + 5y = 1$

$\Rightarrow5y = -5$

$\Rightarrow y = -1$

Hence,

$x=3,\ y=-1$ is the solution of the given system of linear equations.

Solve the following pairs of linear equation:

$\dfrac{2}{x}+\dfrac{2}{3y}=\dfrac{1}{6}$

$\dfrac{2}{x}-\dfrac{1}{y}= 1$

Let

$a=\dfrac 1x,b=\dfrac 1y$

$\Rightarrow 2a+\dfrac 23 b=\dfrac 16.....(i)$ and

$2a-b=1$

$\Rightarrow 2a=1+b$

Put in

$(i)$

$\Rightarrow 1+b+\dfrac 23b=\dfrac 16$

$\Rightarrow 1+\dfrac 53 b=\dfrac 16$

$\Rightarrow\dfrac 53b=-\dfrac 56$

$\Rightarrow b=-\dfrac 12\implies a=\dfrac 14$

$\therefore x=4,y=-2$

Solve the following equation
$3x-4y=1$ ;
$-2x+\dfrac{8}{3}y=5$

The given equations are

$3x-4y=1$ ......(1)

$-2x+\frac{8}{3}=5$ .....(2)

multiply (1) by 2 and (2) by 3

$6x-8y=2$ ....(3)

$-6y+8y=15$ .....(4)

Add equation (3) and (4)

$0=17$ which is not possible Therefore the two equations goes not have any solution.

solve the following equation graphically.
$x+2y=5$ and $3x-y=1$

$x+2y=5$

$3x-y=1$

These lines intersect at the point

$P(1,2)$ and therefore, the solution of given equations is

$x=1, y=2$ .

1337950_1112372_ans_c5eba415911c422fa9694d92506b48cb.PNG

Find the solution of lines : $\begin{array}{l}x + y = 4\\2x - 3y = 3\end{array}$

$x+y=4$

$2x-3y=3$

$x+y=4\,\,\,\,\Rightarrow 2x+2y=8 \Rightarrow 5y=\,5\Rightarrow y=1$

$2x-3y=3\Rightarrow 2x-3y=3\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x=3$

From the figure it is clear that

Sol is

$x=3, y=1$

Sol is

$x=3,y=1$

$\therefore \boxed{(x,y)=(3,1)}$

1085757_1112474_ans_4c49703071d846c8ab45daf0482221c7.png

Solve the following pair of linear equation graphically.
$x + 3y = 6;2x - 3y = 12$

$x+3y=6......(1)$

$2x-3y=12....(2)$

Solving

$(1)$ [Ref image 1]

$4+3y-6=0$

When

$x=0$

$y=2$

When

$y=0$

$x=6$ (Ref image (1)).

Solving

$(2)$ [Ref. image 2]

When

$x=0$

$y=-4$

When

$y=0$

$x=6$

$\therefore$ They intersect at

$(6, 0)$ [Ref. image 3]

1025299_1112228_ans_260fb6c928ce432d8e06805c4dcf4246.png

Solve the following sets of simultaneous equations.
$3x-5y=16; x-3y=8$

Given

$x-3y=8$

$x=8+3y.......(1)$

$3x-5y=16$

substitute value of

$x$ from (1) in the above equation, we get

$3(8+3y)-5y=16$

$24+9y-5y=16$

$9y-5y=16-24$

$4y=-8$

$y=\frac{-8}{4}$

$y=-2$

Putting value of

$y$ in (1) ,we get

$x=8+3y$

$x=8+3\times(-2)$

$x=8-6$

$x=2$

Therefore,

$x=2$ and

$y=-2$

If $3x-2y=5$ and $3y-2x=3$ then find the value of $(x+y)$ .

$3x-2y=5$ ............

$(1)$

$3y-2x=3$

$-2x+3y=3$

Multiply

$(1)$ by

$2$ & eq

$(2)$ by

$3$

$6x-4y=10.......... (3)\\ -6x+9y=19.......... (4)\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_\\ Adding\ eqn\ (3)\ and\ (4),\ we\ get$

$6x-4y=10\\ -6x+9y=19\\ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \\ \quad \quad \quad \quad 5y=1$

$y=\dfrac{1}{5}$

Putting the value in

$6x-\dfrac{4}{5}=10$

$6x=10+\dfrac{4}{5}=\dfrac{54}{5}$

$x=\dfrac{54}{5\times 6}=\dfrac{9}{5}$

$x=\dfrac{9}{5}$

Now

$x+y=\dfrac{9}{5}+\dfrac{1}{5}=\dfrac{10}{5}$ =2$$

$x+y=2$

If $\dfrac{1}{x-y}=\dfrac{1}{2}$ and $\dfrac{1}{x+y}=\dfrac{1}{14}$ then find the value of x.

$\begin{array}{l} \dfrac { 1 }{ { x-y } } =\dfrac { 1 }{ 2 } \, \, \, \, \, \, and\, \, \, \, \dfrac { 1 }{ { x+y } } =\dfrac { 1 }{ { 14 } } \\ x-y=2...\left( i \right) \, \, \, \, and\, \, \, \, x+y=13....\left( { ii } \right) \\ by\, adding\, eq\, \left( i \right) \, \& \left( { ii } \right) \\ \left. { \underline { \, \begin{array}{l} x-y=2 \\ x+y=14 \end{array}\, } } \right| \\ \quad \quad 2x=16 \\ \quad \quad \quad x=8 \end{array}$

Solve:
$\dfrac{5}{x-1}+\dfrac{1}{y-1}=2,\dfrac{6}{x-1}-\dfrac{3}{y-1}=1$

$\dfrac{5}{x-1}+\dfrac{1}{y-1}=2$ .........

$(1)$

$\dfrac{6}{x-1}-\dfrac{3}{y-1}=1$ ...........

$(2)$

Put

$u=\dfrac{1}{x-1},\,v=\dfrac{1}{y-1}$

$\Rightarrow\,5u+v=2$ ........

$(3)$

$\Rightarrow\,6u-3v=1$ ........

$(4)$

$3\times\,equation(3)+1\times\,equation (4)$ we get

$\Rightarrow\,15u+3v+6u-3v=6+1$

$\Rightarrow\,21u=7$

$\Rightarrow\,u=\dfrac{7}{21}=\dfrac{1}{3}$

$\Rightarrow\,\dfrac{1}{x-1}=\dfrac{1}{3}$

$\Rightarrow\,x-1=3$

$\Rightarrow\,x=3+1=4$

Put

$u=\dfrac{1}{3}$ in

$(3)$ we get

$5\times \dfrac{1}{3}+v=2$

$\Rightarrow\,v=2-\dfrac{5}{3}=\dfrac{6-5}{3}=\dfrac{1}{3}$

$\Rightarrow\,\dfrac{1}{y-1}=\dfrac{1}{3}$

$\Rightarrow\,y-1=3$

$\Rightarrow\,y=3+1=4$

$\therefore\,x=4,\,y=4$

Solve the following sets of simultaneous equations.
$x+y=4; 2x-5y=1$

Given

$x+y=4$

$x=4-y.........(1)$

$2x-5y=1$

Putting the value of

$x$ from (1) in the above equation, we get

$2(4-y)-5y=1$

$8-2y-5y=1$

$8-7y=1$

$7y=8-1$

$7y=7$

$y=1$

Putting the value on

$y$ in (1), we get

$x=4-y$

$x=4-1$

$x=3$

Therefore,

$x=3$ and

$y=1$

Solve the following sets of simultaneous equations.
$2x+y=5; 3x-y=5$

Given

$2x+y=5$

$y=5-2x.......(1)$

$3x-y=5$

substitute value of

$y$ from (1) in the above equation, we get

$3x-(5-2x)=5$

$3x-5+2x=5$

$3x+2x=5+5$

$5x=10$

$x=\dfrac{10}{5}$

$x=2$

Putting value of

$x$ in (1) ,we get

$y=5-2x$

$y=5-2\times2$

$y=5-4$

$y=1$

Therefore,

$x=2$ and

$y=1$

Solve the following pairs of liner equation by the method of substitution:
$3x-y=0, x-y+6=0$

In the method of substitution, we find the value of one variable from one equation and then substituent in other. Then solve

Given:-

$3x-y=0$ ......

$1$

$x-y+6=0$ ....

$2$

Using

$1$ ,

$y=3x$ .....

$3$

Put in

$2$ ,

$4x-3x+6=0$

$-2x+6=0$

$2x=6$

$x=3$

Put in

$3$ , to find

$y$

$y=3x$

$y=3\times 3=9$

$y=9$

For what value of $K$ will the following pair of linear equations have infinitely many solutions?

$Kx+3y-(K-3)=0$
$12x+Ky-K=0$

Given

$kx+3y-(k-3)=0$

Comparing with

$a_1x+b_1y+c1=0$

$\therefore a_1=k,\ b_1=3,\ c=-(k-3)$

$12x+ky-k=0$

Comparing with

$a_1x+b_1y+c1=0$

$\therefore a_1=12,\ b_1=k,\ c=-k$

Since equation has infinite number of solutions

So,

$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$

$\dfrac{k}{12}=\dfrac{3}{k}=\dfrac{k-3}{k}$

$\dfrac{k}{12}=\dfrac{3}{k}$

$k^2=12\times3$

$k^2=36$

$k=\pm 6$

$\dfrac{3}{k}=\dfrac{k-3}{k}$

$3k=k(k-3)$

$3k=k^2-3k$

$k^2-3k-3k=0$

$k^2-6k=0$

$k(k-6)=0$

$k=0,6$

Therefore,

$k=6$ satisfies both equations

Hence,

$k=6$

Solve by elimination:
$12x+15y=-18$
$18x-7y=-86$

Given equations

$12x+15y=-18.....(1)$

$18x-7y=-86.......(2)$

Multiplying (1) by

$7$ and (2) by

$15$ ,we get

$84x+105y=-126.....(3)$

$+$

$270x-105y=-1290.....(4)$

-------------------------------------

$84x+270x=-126-1290$

$354x=-1416$

$x=-4$

Putting

$x=-4$ in (1), we get

$12(-4)+15y=-18$

$15y=48-18$

$15y=30$

$y=2$

Solve by elimination
$8x+13y=29$
$12x-7y=17$

Given equations

$8x+13y=29.....(1)$

$12x-7y=17.......(2)$

Multiplying (1) by

$7$ and (2) by

$13$ ,we get

$56x+91y=203.....(3)$

$156x-91y=221.....(4)$

-------------------------------------

$56x+156x=203+221$

$212x=424$

$x=2$

Putting

$x=2$ in (1), we get

$8(2)+13y=29$

$13y=29-16$

$13y=13$

$y=1$

Solve the following sets of simultaneous equations.
$2x+3y+4=0; x-5y=11$

Given

$x-5y=11$

$x=11+5y.......(1)$

$2x+3y+4=0$

substitute value of

$x$ from (1) in the above equation, we get

$2(11+5y)+3y=-4$

$22+10y+3y=-4$

$10y+3y=-4-22$

$13y=-26$

$y=\frac{-26}{13}$

$y=-2$

Putting value of

$y$ in (1) ,we get

$x=11+5y$

$x=11+5\times(-2)$

$x=11-10$

$x=1$

Therefore,

$x=1$ and

$y=-2$

Solve : $\dfrac{3x}{2} - \dfrac{5y }{3} = -2$ , $\dfrac{x}{3} + \dfrac{y}{2} = \dfrac{13}{6}$

$\dfrac{3x}{2}-\dfrac{5y}{3}=-2\Rightarrow 9x-10y=-12$ ...(1)

$\dfrac{x}{3}+\dfrac{y}{2}=\dfrac{13}{6}\Rightarrow 2x+3y=13$ ...(2)

uatiion

Multiplying equation (1) by 3 and (2) by 10,we get

$27x-30y=-36$ ...(3)

$20x+30y=130$ ...(4)

Adding equation (3) and (4),we get

Gives,

$y=3$

Using the value of y in equation (2), we get

$2x+3y-13=0$

$2x+3(3)-13=0$

$2x=13-9=4$

$x=2$

Solve the following sets of simultaneous equations.
$2x-7y=7; 3x+y=22$

Given

$3x+y=22$

$y=22-3x.......(1)$

$2x-7y=7$

substitute value of

$y$ from (1) in the above equation, we get

$2x-7(22-3x)=7$

$2x-154+21x=7$

$2x+21x=7+154$

$23x=161$

$x=\dfrac{161}{23}$

$x=7$

Putting value of

$x$ in (1) ,we get

$y=22-3x$

$y=22-3\times(7)$

$y=22-21$

$y=1$

Therefore,

$x=7$ and

$y=1$

A passenger is allowed some free luggage by Rajdhani express. Two passengers have a total of $105kg$ luggage and they have to pay $Rs. 18$ and $Rs. 12$ as luggage charges respectively. If one passenger having the same $kg$ of luggage then he would have to pay $Rs.36$ , find how much free luggage each passenger is allowed to carry.

Let the free luggage allowed be

$x\ kg$ and the charge paid for each

$kg$ more than

$x=Rs.y$

Then according to the question,

105kgTwo passengers have a total of

$105 kg$ luggage and they have to pay

$Rs.18$ Rs.18 and

$Rs.12$ with some

$x\ kg$ as free luggage .Rs.12

$\therefore \ y(105-(x+x))=12+18$

$105-2x=\dfrac {30}{y}-----(i)$

Also, If one passenger having the same

$kg$ of luggage then he would have to pay

$Rs. 36$ , therefore,

$y(105-x)=36$

$105-x=\dfrac {36}{y}-----(ii)$

Multiplying

$(i)$ by

$36$ and

$(ii)$ by

$30$ we get,

$\dfrac { \begin{matrix} 3780-72x=\dfrac { 1080 }{ y } \\ 3150-30x=\dfrac { 1080 }{ y } \\ -\quad \quad +\quad \quad - \end{matrix} }{ 630-42x=0\\ \quad 42x=630 } \\$

$x=\dfrac {630}{42}=15$

$\therefore \$ Free luggage allowed

$=15\ kg$

From

$(i)$ we get,

$\ 105-2(15)=\dfrac {30}{y}$

$105-30=\dfrac {30}{y}$

$y=\dfrac {30}{75}=0.4$

$y=Rs.0.40=40$ paise

Let us check

$(ii)$

$y(105-x)=30$

$0.4(105-15)=0.4(90)$

$=36$

Solve the following equations by graphical method $2x+3y=0$ .

REF.Image

$2x+3y=0$ solve by graphical method

$2x+3y=0$

$2x=-3y$

$x=-\dfrac{3y}{2}$

when

$y=0, x=0$

$y=2, x=-3$

Now

$y=-2, =x=3$

1370954_1227611_ans_0fc460cf17ea46c7a387acc2d6b77843.JPG

Solve for $x$ and $y$
$3 x - 4 y = 26$
$6 x - 10 y = 40$

$\\3x-4y=26$

multiply this equation by $2$ , we get

$6x-8y=52\>\rightarrow (1)$

$6x-10y=40\>\rightarrow(2)$

$subtracting\>(1) from (2),$

we get

$2y=12$

$y=6$

$then\>3x-4y=26$

$\>3x=26+4y$

$=26+24$

$x=(\dfrac{50}{3})$

The cost of some books is $60$ . Cost of every book is the same. If the book sell decreases the cost of the book by $1$ , one gets $5$ books more for the same amount. What is the cost of each book and how many books are purchased?

let the cost of each book be

$Rs. x$

and no. of books be

$y$

A/c

$xy=60\quad---(1)$

Also

$(x-1)(y+5)=60\quad----(2)$

Solving

$(1)$ and

$(2)$ eqn we get

$x=\dfrac{60}{y}\Longleftrightarrow \left(\dfrac{60}{y}-1\right)(y+5)=60$

$(60-y)(y+5)=60y$

$60y+300-y^2-5y=60y$

$y^2+5y-300=0$

$y=15\quad y=-20$ neglecting

$(y=-20)$

We get no. of books

$=15$

cost of each book

$=\dfrac{60}{15}=Rs. 4$

If $\frac{{5 + 2\sqrt 3 }}{{7 + 4\sqrt 3 }} =$ $a + b\sqrt 3$ , then find the value of a and b

$\begin{array}{l} \frac { { \left( { 5+2\sqrt { 3 } } \right) \left( { 7-4\sqrt { 3 } } \right) } }{ { \left( { 7+4\sqrt { 3 } } \right) \left( { 7-4\sqrt { 3 } } \right) } } =\frac { { \left( { 5+2\sqrt { 3 } } \right) \left( { 7-4\sqrt { 3 } } \right) } }{ { { 7^{ 2 } }-{ { \left( { 4\sqrt { 3 } } \right) }^{ 2 } } } } \\ =11-6\sqrt { 3 } \\ Now,\, compare\, with\, a+b\sqrt { 3 } \\ Therefore, \\ a=11,b=-6 \end{array}$

A student bought some pens at $Rs 8$ each and some pencils at $Rs1.50$ each. If total number of pens and pencils purchased is 16 and their total cost is $Rs 50$ , how many pens did he buy?

$\\Let\>the\>number\>of\>pens\>be\>x\>and\\number\>of\>pencils\>be\>y\\then\\x+y=16\>or\>y=16-x\\and\>their\>value\\8x+1.5y=50\\8x+(\frac{3}{2})(16-x)=50\\8x+24-(\frac{3}{2})x=50\\8x-(\frac{3}{2})x=50-24\\(\frac{13x}{2})=26\>\>\therefore\>x=4\\then\>y=16-x=16-4=12$

Solve the following pair of linear equations :
$0.2x+0.3y=1.3$
$0.4x+0.5y=2.3$

$0.2x+0.3y=1.3 ......(1)$

$0.4x+0.5y=1.3 ......(2)$

Derive the value of

$y$ from equation (1)

$y = \dfrac{{1.3 - 0.2x}}{{0.3}}$ ......(3)

Substitute value of

$y$ in equation(3)

$0.4x + 0.5\left( {\dfrac{{1.3 - 0.2x}}{{0.3}}} \right) = 2.3$

$\Rightarrow$

$0.12x + 0.5\left( {1.3 - 0.2x} \right) = 0.69$

$\Rightarrow$

$0.12x + 0.65 - 0.1x = 0.69$

$\Rightarrow$

$0.02x = 0.64$

$\Rightarrow$

$x = 32$

Substitute value of

$x$ in equation (3)

$y = \dfrac{{1.3 - 0.2\left( {32} \right)}}{{0.3}}$

$= \dfrac{{1.3 - 6.4}}{{0.3}}$

$= - 17$

So,

$x=32$ and

$y=-17$

Solve
$x+y=3$
$x-y=1$

$x+y=3$

$x-y=1$

$2x=4\rightarrow x=2 \rightarrow y=1$

Solve the following equations.
$3x-y=7; x+4y=11$

$3x-7=7....(1)$

$x+4y=11....(2)$

Multiply (1) by 4

$12x-4y=28....(3)$

$x+4y=11....(2)$

Add (3) and (2)

$13x=39$

$x=3$

$a-y=7$

$y=2$

Draw the graphs of the pair of linear equations in two variables $x + 3 y = 6,2 x - y = 5$ . Find its solution set.

$x + 3y = 6$

$+ (2x - y = 5)3$

______________

$7x = 21$

______________

$x = 3 \, ; \, y = 1$

writing equation in intercept form

$\dfrac{x}{6} + \dfrac{y}{2} = 1$

$\dfrac{x}{\dfrac{5}{2}} + \dfrac{y}{-5} = 1$

1330428_1227646_ans_7fb221a5d47d4e3298de0710dd694396.png

Solve: $x+2y=3,\,\,2x-y=5$

$x+2y=3$

$\Rightarrow x=3-2y$

Substituting

$x=3-2y$ in

$2x-y=5$ ,

$\begin{aligned}{}2\left( {3 - 2y} \right) - y &= 5\\6 - 4y - y& = 5\\ - 5y &= 5 - 6\\ - 5y& = - 1\\y &= \frac{1}{5}\end{aligned}$

Substitute

$y=\dfrac{1}{5}$ in

$x=3-2y$ we get,

$\begin{aligned}{}x &= 3 - 2 \times \frac{1}{5}\\ &= \frac{{15 - 2}}{5}\\ &= \frac{{13}}{5}\end{aligned}$

$\therefore \,\,x=\dfrac{13}{5},\,y=\dfrac{1}{5}$

What value of K, will the following pair of the linear equations have infinitely many solutions?
$2x+3y-13=0 ; 6x-ky-39=0$

We have,

$2x+3y-13=0$

$..........(1)$

$6x-ky-39=0$

$..........(2)$

Since, linear equations have infinitely many solution.

$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$

Therefore,

$\dfrac{2}{6}=\dfrac{3}{-k}=\dfrac{-13}{-39}$

$\dfrac{1}{3}=\dfrac{3}{-k}$

$\therefore k=-9$

Find the value of x and y

$\dfrac{5}{{x - 1}} + \dfrac{1}{{y - 2}} = 2$ and $\dfrac{6}{{x - 1}} - \dfrac{3}{{y - 2}} = 1$

$\dfrac{5}{x-1}+\dfrac{1}{y-2}=2\,\,\,\,\,\dfrac{6}{x-1}-\dfrac{3}{y-2}=1$

Let

$\dfrac{1}{x-1}=x\,\,, \dfrac{1}{y-2}=y$

$5x+y=2$ .......(1)

$6x-3y=1$ ........(2)

$3(1)+(2)$

$\Rightarrow 15x+3y=6$

$+ 6x-3y=1$

$\overline{21x=7\Rightarrow }x=\dfrac{1}{3},y=\dfrac{1}{3}$

$\dfrac{1}{x-1}=\dfrac{1}{3},\dfrac{1}{y-2}=\dfrac{1}{3}$

$x=4$

$y=5$

Solve the pair of linear equation $: 2X-5Y-4=0, 4X-8Y=8.$

Given first equation is

$2X-5Y =4\, ...(1)$

Second equation,

$4X-8Y=8$

$\Rightarrow X-2Y=2$

$\Rightarrow X=2+2Y$

From equation (1)

$2(2+2Y)-5Y=4$

$4+4Y-5Y=4$

$4-Y=4$

$4-4=Y$

So,

$Y=0$

Substituting the value of

$Y$ in (1),

$2X-5(0)=4$

$2X=4$

$X=2$

Solve :
$x+5y=17\\2x+3y=13$

$x+5y=17$

$x=17-5y$

$2x+3y=13$

$2(17-5y)+3y=13$

$34-10y+3y=13$

$-7y=-21$

$y=3$

$x=17-5(3)$

$x=17-15=2$

$x=2;y=3$

Solve $x+4y=24xy$ , $x+2y=21xy$ .

$x+4y=24xy\quad \longrightarrow \left( I \right)$

$x+2y=21xy\quad \longrightarrow \left( II \right)$

Divide both equation by

$xy$

$\dfrac { 1 }{ y } +\dfrac { 4 }{ x } =24\quad \longrightarrow \left( III \right)$

$\dfrac { 1 }{ y } +\dfrac { 2 }{ x } =21\quad \longrightarrow \left( IV \right)$

Subtraction

$(IV)$ from

$(III)$

$\dfrac { 2 }{ x } =3$

$x=\dfrac { 2 }{ 3 }$

From equation

$(IV)$

So,

$\dfrac { 1 }{ y } +3=21$

$y=\dfrac { 1 }{ 18 }$

$x=\dfrac { 2 }{ 3 }$ and

$y=\dfrac { 1 }{ 18 }$

Solve the following pair of an equation by elimination the coefficient method:
$x+2y=11$
$2x-y=2$

Consider given the equations,

$x+2y=11$ ……..(1)

$2x-y=2$ ………(2)

Now, multiply by 2 in equation 2^nd,

$4x-2y=4\,\,\,\,\,\,\,....(3)$

$x+2y=11\,\,\,\,\,\,.......(1)$

$5x=15$ (add equation 1st and 3rd)

$x=3$

Put, the value of x in equation 1^st,

$3+2\times y=11$

$2y=8$

$y=4$

Hence, this is the answer.

Solve the following pair of linear equation by substitution method:
$3x + 2y - 7 = 0$
$4x + y - 6 = 0$

$3x+2y=7$

$4x+y=6\Rightarrow y=6-4x$

placing in equation (1)

$3x+2(6 - 4x)=7$

$3x+12-8x=7 \Rightarrow -5x+12=7$

$-5x=7-12\Rightarrow -5x=-5$

x=1

$\therefore$ y = 6 - 4

$\times 1 = 2$

$\therefore$ x = 1, y = 2

1126262_1292746_ans_e0bd629e2ed048e08ee69da04887dd64.JPG

Find the equations of the lines for which $\tan{\theta}=\dfrac{3}{4}$ , where $\theta$ is the angle of inclination of the line and $y-$ intercept is $\dfrac{-5}{3}$

Slope of the line is

$m=\tan{\theta}=\dfrac{3}{4}$

Using slope-intercept form, the equation is

$y=mx+c$

$y=\dfrac{3}{4}x+\dfrac{-5}{3}$

$\Rightarrow\,12y=9x-20$

$\therefore\,$ equation of the line is

$9x-12y-20=0$

Simplify:
$0.2x + 0.3y = 1.3$
$0.4x + 0.5y = 2.3$

$0.2x+0.3y = 1.3$

$(0.2x+0.3y)10=1.3\times 10$

$\Rightarrow 2x+3y=13...(1)$

$0.4x+0.5y=2.3$

$(0.4x+0.5y)5=2.3\times 5$

$2x+2.5y=11.5...(2)$

Equation

$(1)$ &

$(2)$

$2 x+3 y-2 x-2.5 y=13-11.5$ .

$\Rightarrow 0.5 y = 1.5$

$\Rightarrow 5 y = 15$

$\Rightarrow y = \dfrac {15}{5} = 3$

$x = (13-\dfrac {3y}{2})$

$= (13-\dfrac {3.3}{2})$

$\Rightarrow$

$\dfrac{13-9}{2 }=\dfrac 42 = 2$

$\therefore x = 2;y = 3$

Solve the following system of equation
$2x +3y +8 = 0$
$4x + 5y + 14 = 0$

Given,

$2x+3y+8 = 0$ ...(1)

$4x+5y+14 = 0$ ...(2)

multiply equation. (1) by 5 and equation (2) by 3 we get

$10x + 15y + 40 = 0$

$12x + 15y + 42 = 0$

$- - -$

___________________

$-2x-2=0$

$[x = -1]$ from equation.(1)

$3y = 8 -2x$

$y = \dfrac{10}{3}$

Hence using cross multiplication method intersection point of line 1 & 2 are

$(-1,\dfrac{10}{3} )$

Solve:
$x+y=7;x-y=3$

$x+y=7\cdots(1)\\x-y=3\\x=y+3$

Put it in

$(1)$

$y+3+y=7\\2y+3=7\\2y=7-3\\2y=4\\y=\dfrac{4}{2}=2$

Put

$y=2$ in

$x+y=7$ we get

$x=7-y=7-2=5$

$\therefore x=5,y=2$

Solve :
$3x-2y=1$
$2x+y=3$

Given: two linear equation:

$3x-2y=1$ ........(1)

$2x+y=3$ ...... (2)

It gives:

$y=3-2x$

Put the value of

$y$ in (1) we get:

$\Rightarrow 3x-2(3-2x)=1$

$\Rightarrow3x-6+4x=1$

$\Rightarrow7x=7$

$\Rightarrow x=1$

Put the above value of

$x$ in eq.(1) we get:

$3(1)-2y=1$

$\Rightarrow 1-2y=1$

$\Rightarrow2y=2$

$y=1$

So, the solution is

$(1,1)$ .

Solve:
$x-y=5;3x+2y=25$

$x-y=5---(1)$

$3x+2y= 25---(2)$

$3x-3y=25$

$3x+2y=25$

_______________

$+5 y = +10$

$\Rightarrow 5y=10$

$\Rightarrow y=\dfrac {10}{5}$

$\Rightarrow y=2.$

sub

$y = 2$ in

$(1)$

$x-y = 5$

$\Rightarrow x-2=5$

$\Rightarrow x=5+2$

$\Rightarrow x=7$

$\therefore$

$x = 7$ and

$y = 2$ .

Solve the following simultaneous equation graphically.
$2x+3y=12;\ x-y=1$

$2x + 3y = 12$

$y=\dfrac{12-2x}{3}$

Put

$x=0$ ,

$y=\dfrac{12}{3}$

$\Rightarrow y=4$

Put

$x=6$

$\dfrac{0}{3}$

$\Rightarrow y=0$

Draw the line joining the point

$(0,4)$ and

$(6,0)$ to get the graph of

$2x+3y=12$

Consider,

$x - y = 1$

$y=x-1$

Put

$x=0$

$\Rightarrow y=-1$

Put

$x=1$

$\Rightarrow y=0$

Draw the line joining the point

$(0,-1)$ and

$(1,0)$ to get the graph of

$x-y=1$

From the graph it can be seen that point of intersection of two lines

$(3,2)$

Hence required solution is

$(3,2)$

1812028_1370221_ans_9b42d5416f29442985548db973ffc5a5.png

Solve the following pairs of equations by reducing them to a pair of linear equations.
$\dfrac{5}{x-1}-\dfrac{1}{y-2}=2$ & $\dfrac{6}{x-1}-\dfrac{3}{y-2}=1$

$\dfrac{5}{x-1}-\dfrac{1}{y-2}=2$ .....

$(1)$

$\dfrac{6}{x-1}-\dfrac{3}{y-2}=1$ .....

$(2)$

Let

$u=\dfrac{1}{x-1}$ and

$v=\dfrac{1}{y-2}$

then the above equations

$(1)$ and

$(2)$ becomes

$5u-v=2$ .....

$(3)$

$6u-3v=1$ .....

$(4)$

$6\times$ eqn

$(3)-5\times$ eqn

$(4)$ gives

$30u-6v-30u+15v=12-5$

$\Rightarrow 9v=7$

$\Rightarrow v=\dfrac{9}{7}$ in eqn

$(3)$ we get

Since

$v=\dfrac{1}{y-2}\Rightarrow y-2=\dfrac{7}{9}$

$\therefore y=\dfrac{7}{9}+2=\dfrac{7+18}{9}=\dfrac{25}{9}$

Substitute

$v=\dfrac{9}{7}$ in

$5u-v=2$

$\Rightarrow 5u=2+v=2+\dfrac{9}{7}=\dfrac{14+9}{7}=\dfrac{23}{7}$

$\Rightarrow u=\dfrac{23}{35}$

$\Rightarrow \dfrac{1}{x-1}=\dfrac{23}{35}$

$\Rightarrow x-1=\dfrac{35}{23}$

$\Rightarrow 23x-23=35$

$\Rightarrow 23x=35+23=58$

$\therefore x=\dfrac{58}{23}$

Hence

$x=\dfrac{58}{23},y=\dfrac{25}{9}$

Solve: $3 x - 4 y = -7$ ; $5 x - 2 y = 0$

$3x-4y=-7 ... (1)$

$5x-2y=0 ... (2)$

$5x=2y$

$x= \dfrac{2y}{5}$

$3(\dfrac{2y}{5}) - 4y=-7$

$(By \,substituting\, x=\dfrac{2y}{5}\, in\, equation ...(1))$

$\dfrac{6y}{5} - 4y = -7$

$\dfrac{-14}{5} y = -y$

$y=7 \times \dfrac{5}{14} = \dfrac{5}{2}$

$5x-2\times \dfrac{5}{2} = 0$

$(By\,\, substituting\,\, y= \dfrac{5}{2} \,\,in \,\,equation ...(2))$

$5x-5=0$

$5x=5$

$x=1$

Solve:
$\dfrac { x+1 }{ 2 } +\dfrac { y-1 }{ 3 } =8$

$\dfrac { x-1 }{ 3 } +\dfrac { y+1 }{ 2 } =9$

Given:

$\dfrac{x+1}{2}+\dfrac{y-1}{3}=8$

$\dfrac{3x+3}{2\times 3}+\dfrac{2y-2}{3\times 2}=8$

$\dfrac{3x+3}{6}+\dfrac{2y-2}{6}=8$

$\dfrac{3x+3+2y-2}{6}=8$

$3x+2y+1=48$

$3x+2y=47------(1)$

$\dfrac{x-1}{3}+\dfrac{y+1}{2}=9$

$\dfrac{2x-2}{2\times 3}+\dfrac{3y+3}{3\times 2}=9$

$\dfrac{2x-2}{6}+\dfrac{3y+3}{6}=9$

$\dfrac{2x-2+3y+3}{6}=9$

$2x+3y+1=54$

$2x+3y=53$ -------- (2)

By solving (1) and (2)

$3x+2y=47-------\times 3$

$2x+3y=53-------\times 2$

$9x+6y=141$

$(-)$

$4x+6y=106$

$5x=35$

$x=7$

Substituting the value of

$x$ in

$(1)$

$3(7)+2y=47$

$21+2y=47$

$2y=47-21$

$2y=26$

$y=13$

So,

$x=7$ and

$y=13$

If $12x +17y = 48$ and $17x +12y = 68$ , then find $x-y$ .

Given

$12x+17y=48.....\times 12$

$17x+12y=68.....\times 17$

$144x+204y=576$

$289x+204y=1156$

$-145x=-580$

$x=4$

substitute

$x$ value in the 1st equation

$12\times 4+17y=48$

$48+17y=48$

$17y=0$

then

$x-y=4-0=4$

Solve the following equation:
$\cfrac{3x}{2} +\cfrac {2y}{3} = \cfrac{40}{3}$ ; $\cfrac {2x}{3} +\cfrac {3y}{2} = \cfrac{25}{3}$

We have

$\cfrac{3x}{2} +\cfrac {2y}{3} = \cfrac{40}{3}$ ;

$\cfrac {2x}{3} +\cfrac {3y}{2} = \cfrac{25}{3}$

Multiply

$2/3$ in first and

$3/2$ in second

$x+\dfrac{4y}{9}=\dfrac{80}{9}$ .............(i) and

$x+\dfrac{9y}{4}=\dfrac{25}{2}$ .............(ii)

Substract (i) from (ii)

$\dfrac{9y}{4}-\dfrac{4y}{9}=\dfrac{25}{2}-\dfrac{80}{9}\Rightarrow \dfrac{81y-16y}{36}=\dfrac{225-160}{18}$

$\Rightarrow \dfrac{65y}{36}=\dfrac{65}{18}$

putting

$y=2$ in (1) we get

$y=2$

$x+\dfrac{8}{9}=\dfrac{80}{9}\Rightarrow x=8$

$x=8$ and

$y=2$

Solve the equations graphically .
$x+y=10$
$y=x+4$

$x+y=10$

$x+x+4=10$ since

$y=x+4$

$2x=10-4=6$

$x=\dfrac{6}{2}=3$

Put

$x=3$ in

$y=x+4$ we get

$y=3+4=7$

$\therefore \left(3,7\right)$ is the point of intersection of the lines

$x+y=4$ and

$y=x+4$ as shown in the graph.

1211325_1505716_ans_bc841941530e4cf7b6cfc499a1fb359a.png

Solve the following simultaneous equation
$\dfrac{{30}}{{x - y}} + \dfrac{{44}}{{x + y}} = 10\ ;\\ \dfrac{{40}}{{x - y}} + \dfrac{{55}}{{x + y}} = 13$

Given equations are,

$\dfrac{30}{x-y}+\dfrac{44}{x+y}=10$ ...(i)

$\dfrac{40}{x-y}+\dfrac{55}{x+y}=13$ ...(ii)

Multiplying eq.(i) by 4 and eq(ii) by 3, we get

$\dfrac{120}{x-y}+\dfrac{176}{x+y}=40$ ...(iii)

$\dfrac{120}{x-y}+\dfrac{165}{x+y}=39$ ...(iv)

Subtracting eq.(iv) from eq.(iii), we get

$0+\dfrac{11}{x+y}=1$

$\Rightarrow x+y=11$ ...(v)

Substituting eq.(v) in eq.(i) , we get

$\dfrac{30}{x-y}+\dfrac{44}{11}=10$

$\Rightarrow \dfrac{30}{x-y}=10-4$

$\Rightarrow \dfrac{30}{x-y}=6$

$\Rightarrow x-y=\dfrac{30}{6}$

$\Rightarrow x-y=5$ ...(vi)

Adding eq.(v) and eq.(vi), we get

$2x=16$

$\therefore x=16/2=8$

Substituting the value of

$x$ in eq.(v)

$8+y=11$

$\Rightarrow y=11-8$

$\therefore y=3$

Hence,

$x=8$ and

$y=3$

Solve the simultaneously equations $2y - x = 0, 10x+15y=105$

The given equations can be solved by substitution method.

$2y-x=0$

$2y=x$

$\therefore x=2y$

$10x+15y=105$ ………….

$(1)$

Substitute

$x$ value in

$(1)$

$10x+15y=105$

$\Rightarrow 10(2y)+15y=105$

$\Rightarrow 20y+15y=105$

$\Rightarrow 35y=105$

$\Rightarrow y=\dfrac {105}{35}$

$\Rightarrow y=3$

Sub y

$=3$ in

$x$

$x=2y$

$\Rightarrow x=2(3)$

$\Rightarrow x=6$

$\therefore x=6$ and

$y=3$ .

Solve : $10x+3y=75$ ; $6x-5y=11$

$\textbf{Hint: Use substitution method}$

Solution:

$\textbf{Step1: Write the value of one variable in terms of other variable.}$

$\quad\quad\quad$ We have given,

$10x + 3y = 75$ …. (1)

$6x - 5y = 11$ …. (2)

$\quad\quad\quad$ Simplify the eq(2), to get the value of variable $x$ in terms of variable $y$ .

$\quad\quad\quad$ $\Rightarrow 6x = 5y + 11$

$\Rightarrow x = \dfrac{{5y + 11}}{6}$ …. (3)

$\textbf{Step 2: Put the value of the variable obtained from the equation in other equation}$

$\quad\quad\quad$ Put the value of $x$ from eq(3) in eq(1), we get

$10\dfrac{{\left( {5y + 11} \right)}}{6} + 3y = 75$

$\Rightarrow \dfrac{{25y + 55}}{3} + 3y = 75$

$\Rightarrow 24y + 55 + 9y = 225$

$\Rightarrow 34y = 170$

$\quad\quad\quad$ $\Rightarrow y = 5$

$\quad\quad\quad$ Substituting the value of $y$ in eq(3), we get

$x = \dfrac{{5 \times 5 + 11}}{6}$

$\Rightarrow x = \dfrac{{36}}{6} = 6$

$\textbf{Hence, the solution of pairs of equation is}$ $x = 6,\;y = 5$ .

Solve by the method of substitution $x+y=5,$ $x-y=1$

Substitution method

$x+y=5$

$x-y=1$

$x=5-y ...(1)$

$x-y=1 ...(2)$

Substituting

$...(1) \,\, in\,\, (2)$

$5-y-y=1$

$5-2y=1$

$2y=5-1$

$2y=4$

$y=2$

Substituting

$y=2 \,\,in\,\, (2)$

$x-y = 1$

$x-2 = 1$

$x=3$

$\therefore x=3$ &

$y=2$

Solve the following pair of linear equations by the elimination method and the subsitution method:
$3x-5y-4=0$ and $9x=2y+7$

$3(3x-5y=4) \Rightarrow 9x-15y=12$

$9x-2y=7$

$\Rightarrow 9x-2y=7$

___________

$- 13y = 5$

$y=\dfrac {-5}{13}$

$x= \dfrac{-10/13 + 7}{9} = \dfrac {9}{13}$

Substitution method

$3x = 5y + 4$

$3.(5y+4 )= 2y+7$

$\Rightarrow 15y + 12 = 2y + 7$

$\Rightarrow 13y=-5 \Rightarrow y=\dfrac{-5}{13}$

$x= \dfrac{-10/13 + 7}{9} = \dfrac {9}{13}$

Solve each of the following system of equations graphically:
$2x+3y=8$ ,
$x-2y+3=0$

$2x+3y=8$ ....(1)

$x-2y+3=0$ ......(2)

Isolate

$x$ from equation (1) and find the value of

$x$ and

$y$ .

$2x+3y=8$

$2x=8-3y$

$x:$	$4$	$1$	$-2$
$y:$	$0$	$2$	$4$

Similarly, isolate

$x$ from equation (2) and find the values of

$x$ and

$y$ .

$x-2y+3=0$
or

$x=2y-3$

$x:$	$-3$	$-1$	$1$
$y:$	$0$	$1$	$2$

Graph:
Both the lines intersect each other at point

$(1, 2)$ .
So,

$x=1, y=2$

1601487_1713734_ans_a5445d54971f440b87d73bf7995c5371.PNG

Solve :
$3x-2y=4$
$2x+y=5$

$3x-2y=4\\2x+y=5\\y=5-2x\\3x-2(5-2x)=4\\3x-10+4x=4\\7x=14\\x=2\\3(2)-2y=4\\6-2y=4\\2y=2\\y=1$

Solve for $x$ and $y$ :
$x-y=3,$
$\dfrac x3+\dfrac y2=6$

$x-y=3 ....(1)$

$\dfrac x3+\dfrac y2=6....(2)$

Find the value of

$x$ from equation and substitute in equation

$(2)$ .

$x=3+y$

Now

$\dfrac {3+y}{3}+\dfrac {y}{2}=6$

$\dfrac {6+2y+3y}{6}=6$

$\Rightarrow \dfrac {6+5y}{6}=6$

$\Rightarrow 6+5y=36\Rightarrow 5y=36-6=30$

$y=\dfrac {30}{5}=6$

and

$x=3+y=3+6=9$

The value of

$x$ is

$9$ and the value of

$y$ is

$6$ .

Solve : $s-t=3$
$\cfrac{s}{3}+\cfrac{t}{2}=6$

$s-t=3..........(i)$

$\cfrac { s }{ 3 } +\cfrac { t }{ 2 } =6$

$2s+3t=36.........(ii)$

Multiply equation

$(i)$ by

$3$

$3s-3t=9............(iii)$

Add equation

$(ii)$ and

$(iii)$

$5s=45$ ,

$s=\dfrac{45}{5}\\\ \ \ =9$

$t=s-3$

$t=9-3\\=6$

$\therefore s=9,t=6$

Solve each of the following system of equations graphically:
$2x+3y=2$
$x-2y=8$

$2x+3y=2$ ......(1)

$x-2y=8$ ........(2)

Step 1: From equation (1), isolate

$x$

$2x=2-3y$
or

$x=(2-3y)/2$

Choose any values for

$y$ and get corresponding value of

$x$ :

$x:$	$1$	$-2$	$4$
$y:$	$0$	$2$	$-2$

Plot above points on the graph and join them.

Step 2: From equation (2)

$x-2y=8$
or

$x=8+2y$
Choose any values for

$y$ and get corresponding value of

$x$ :

$x:$	$6$	$4$	$2$
$y:$	$-1$	$-2$	$-3$

Plot above points on the graph and join them.

Step 3:
Graph:
Step 4: Both the lines intersect each other at point

$(4, -2)$ .
So,

$x=4, y=-2$

1601424_1713731_ans_1a8ddaf01b6b4fc3860c181d7de7cf07.PNG

Solve for $x$ and $y$ :
$x+y=3$ .
$4x-3y=26$ .

$x+y=3 .....(1)$

$4x-3y=26 ....(2)$

Isolate

$x$ from equation

$(1)$ , we get

$x=3-y.....(3)$

Substituting the value of

$x$ in equation

$(2)$ .

$4(3-y)-3y=26$

$or,12-4y-3y=26$

$or,-7y=14$

$or,y=-2$

Now for

$x$ , putting in

$(3)$

$x=3-y=3-(-2)=5$

Answer :

$x=5$ and

$y=-2$

Solve each of the following system of equations graphically:
$3x+2y=4$
$2x-3y=7$

$3x+2y=4$ ....(1)

$2x-3y=7$ ......(2)

Isolate

$x$ from equation (1) and find the value of

$x$ and

$y$ .

$3x+2y=4$
or

$x=(4-2y)/3$

$x:$	$0$	$2$	$4$
$y:$	$2$	$-1$	$-4$

Similarly, isolate

$x$ from equation (2) and find the values of

$x$ and

$y$ .

$2x-3y=7$

$x=(3y+7)/2$

$x:$	$5$	$2$	$-1$
$y:$	$1$	$-1$	$-3$

Graph:
Both the lines intersect each other at point

$(2, -1)$ .
So,

$x=2, y=-1$

1601451_1713733_ans_a07564e33720485586ba6df0f99449a0.PNG

Solve the following pair of linear equations
(i) $x+y=14$
$x-y=4$
(ii) $3x-y=3$
$9x-3y=9$

1308978_1585977_ans_1d6cc526f146448f958412d88a24571a.jpg

Solve each of the following system of equations graphically:
$2x-5y+4=0$ ,
$2x+y-8=0$

$2x-5y+4=0$ ....(1)

$2x+y-8=0$ ......(2)

Isolate

$x$ from equation (1) and find the value of

$x$ and

$y$ .

$2x-5y+4=0$
or

$2x=5y-4$

$x=(5y-4)/2$

$x:$	$-2$	$3$	$8$
$y:$	$0$	$2$	$4$

Similarly, isolate

$x$ from equation (2) and find the values of

$x$ and

$y$ .

$2x+y-8=0$
or

$y=8-2x$

$x:$	$1$	$2$	$3$
$y:$	$6$	$4$	$2$

Graph:
Both the lines intersect each other at point

$(3, 2)$ .
So,

$x=3, y=2$

1601509_1713735_ans_4fcf17ed29cc40c3aca1a6f2a8b1ff04.PNG

Solve each of the following system of equations graphically:
$x+2y+2=0$
$3x+2y-2=0$

$x+2y+2=0$ ....(1)

$3x+2y-2=0$ ......(2)

Isolate

$x$ from equation (1) and find the value of

$x$ and

$y$ .

$x+2y+2=0$
or

$x=-2y-2$

$x:$	$-2$	$0$	$2$
$y:$	$0$	$-1$	$-2$

Similarly, isolate

$x$ from equation (2) and find the values of

$x$ and

$y$ .

$3x+2y-2=0$
or

$x=(2-2y)/3$

$x:$	$0$	$2$	$-2$
$y:$	$1$	$-2$	$4$

Graph:

As expressed in above graph where red line=>

$x+2y+2=0$ & blue line =>

$3x+2y-2=0$

Both the lines intersect each other at point

$(2, -2)$ .
So,

$x=2, y=-2$

1602337_1713767_ans_b98a5f5a008b4fd19a64638388132dfa.PNG

Solve each of the following system of equations graphically:
$2x+3y+5=0$
$3x-2y-12=0$

$2x+3y+5=0$ ....(1)

$3x-2y-12=0$ ......(2)

Isolate

$x$ from equation (1) and find the value of

$x$ and

$y$ .

$2x+3y+5=0$

$2x=-3y-5$

$x=(-3y-5)/2$

$x:$	$-4$	$-1$	$2$
$y:$	$1$	$-1$	$-3$

Similarly, isolate

$x$ from equation (2) and find the values of

$x$ and

$y$ .

$3x-2y-12=0$
or

$3x=2y+12$
or

$x=(2y+12)/3$

$x:$	$4$	$0$	$2$
$y:$	$0$	$-6$	$-3$

Graph:
Both the lines intersect each other at point

$(2, -3)$ .
So,

$x=2, y=-3$

1601748_1713743_ans_33fe8390416142eab94fdb122fe600a9.png

Solve for $x$ and $y$ :
$2x-\dfrac {3y}4=3$ ,
$5x=2y+7$ .

The given equation are

$2x-\dfrac {3y}4=3$ and $5x=2y+7$

Rearranging these equations we get,

$8x-3y=12$ .....(1)

$5x-2y=7$ ......(2)

multiply (1) by 2 and by 3 on subtracting we get

$16x-6y=24$

$15x-6y=21$

- + -

$\overline{x\ \ - \ 0 =\ \ 3}$

$x = 3$

putting $x=3$ in (1) we get

$8(3)-3y=12$

$24-3y=12$

$-3y=12-24$

$-3y=-12$

$y=-12/-3$

$y=4$

hence $x=3,y=4$

Solve each of the following system of equations graphically:
$3x+2y=12$
$5x-2y=4$

$3x+2y=12$ ....(1)

$5x-2y=4$ ......(2)

Isolate

$x$ from equation (1) and find the value of

$x$ and

$y$ .

$3x+2y=12$

$3x=12-2y$

$x:$	$4$	$2$	$0$
$y:$	$0$	$3$	$6$

Similarly, isolate

$x$ from equation (2) and find the values of

$x$ and

$y$ .

$5x-2y=4$

$5x=4+2y$
or

$x=(4+2y)/5$

$x:$	$0$	$2$	$4$
$y:$	$-2$	$3$	$8$

Graph:
Both the lines intersect each other at point

$(2, 3)$ .
So,

$x=2, y=3$

1601623_1713738_ans_311bbd7287024866843fc61e6624f1eb.png

Solve for $x$ and $y$ :
$2x+5y=8/3$ ,
$3x-2y=5/6$ .

$2x+5y=8/3 ...(1)$ ,

$3x-2y=5/6 ...(2)$ .

Let us use elimination method to solve the given system of equation.
Multiply

$(1)$ by

$4$ and

$(2)$ by

$5$ . And add both the equations.

Adding we get

$\dfrac { 24x+60y=32\\ 90x-60y=25 }{ 114x\quad =57 }$

$x=1/2$

Substitute the value of

$x$ in equation

$(1)$ , we have

$2(1/2)+5y=8/3$

$y=1/3$

Answer :

$x=1/2$ and

$y=1/3$

Solve each of the following system of equations graphically:
$2x+3y-4=0$
$3x-y+5=0$

$2x+3y-4=0$ ....(1)

$3x-y+5=0$ ......(2)

Isolate

$x$ from equation (1) and find the values of

$x$ and

$y$ .

$2x+3y-4=0$
or

$x=\dfrac{4-3y}{2}$

Hence,

$x:$	$2$	$-1$	$5$
$y:$	$0$	$2$	$-2$

Now, plot the points

$(2,0),\ (-1,2)\ and\ (5,-2)$ on the graph to get the straight line represented by the equation

$2x+3y-4=0$ .

Similarly, isolate

$x$ from equation (2) and find the values of

$x$ and

$y$ .

$3x-y+5=0$
or

$y=5+3x$

$x:$	$0$	$-1$	$-2$
$y:$	$5$	$2$	$-1$

Now, plot the points

$(0,5),\ (-1,2)\ and\ (-2,-1)$ on the graph to get the straight line represented by the equation

$3x-y+5=0$ .

The graph of both the lines will be as shown in the figure given above.

We can see from the graph that, both the lines intersect each other at point

$(-1, 2)$ .

So,

$x=-1, y=2$ is the solution of the given system of equations.

Hence,

$(-1,2)$ is the required solution.

1601885_1713758_ans_78fc67bb556545feac0b0e875451f406.png

Solve for $x$ and $y$ :
$2x-3y=13$
$7x-2y=20$ .

$2x-3y=13 \quad\quad\dots(1)$

$7x-2y=20 \quad\quad\dots(2)$

Let us use the elimination method for the given system of equations.

Multiply

$(1)$ by

$2$ and

$(2)$ by

$-3$ . And add both the equations.

$\begin{array}{l}\underline {\begin{array}{ccccccccccccccc}{4x}& - &{6y}& = &{26}\\{ - 21x}& + &{6y}& = &{ - 60}\end{array}} \\\;\;\begin{array}{ccccccccccccccc}{ - 17x}&{\;}&{\;}&\;\;\; = &{\; - 34}\end{array}\\\quad\quad\quad\quad\quad\quad\;\;x = 2\end{array}$

Substitute the value of

$x$ in equation

$(1)$ ,

$\begin{aligned}{}2(2) - 3y &= 13\\4 - 3y &= 13\\ - 3y &= 9\\y &= - 3\end{aligned}$

Hence, the value of

$x$ is equal to

$2$ and that of

$y$ is equal to

$-3$ .

Solve each of the following system of equations graphically:
$3x+y+1=0$
$2x-3y+8=0$

$3x+y+1=0$ ....(1)

$2x-3y+8=0$ ......(2)

Isolate

$x$ from equation (1) and find the value of

$x$ and

$y$ .

$3x+y+1=0$
or

$y=-3x-1$

$x:$	$0$	$-1$	$-2$
$y:$	$-1$	$2$	$5$

Similarly, isolate

$x$ from equation (2) and find the values of

$x$ and

$y$ .

$2x-3y+8=0$

$x=(3y-8)/2$

$x:$	$-4$	$-1$	$2$
$y:$	$0$	$2$	$4$

Graph:
Both the lines intersect each other at point

$(-1, 2)$ .
So,

$x=-1, y=2$

1601698_1713739_ans_21d6b092c0b546dfad6f6d4a3d761002.png

Solve each of the following system of equations graphically:
$2x-3y+13=0$
$3x-2y+12=0$

$2x-3y+13=0$ ....(1)

$3x-2y+12=0$ ......(2)

Isolate

$x$ from equation (1) and find the value of

$x$ and

$y$ .

$2x-3y+13=0$
or

$2x=3y-13$
or

$x=(3y-13)/2$

$x:$	$-5$	$-2$	$1$
$y:$	$1$	$3$	$5$

Similarly, isolate

$x$ from equation (2) and find the values of

$x$ and

$y$ .

$3x-2y+12=0$
or

$x=(2y-12)/3$

$x:$	$-4$	$-2$	$0$
$y:$	$0$	$3$	$6$

Graph:
Both the lines intersect each other at point

$(-2, 3)$ .
So,

$x=-2, y=3$

1601842_1713746_ans_e5f69295ce5641c7b5465f7646071ea6.png

Solve for $x$ and $y$ :
$4x-3y=8$
$6x-y=29/3$

$4x-3y=8 ...(1)$

$6x-y=29/3 ...(2)$

Let us use elimination method to solve the given system of equation.
Multiply

$(2)$ by

$3$ . And subtract both the equations.

$\dfrac { 4x-3y=8\\ 18x-3y=29\\ -\quad +\quad - }{ -14x\quad =-21 }$

$x=3/2$

From

$(1)$ ;

$4(3/2)-3y=8$
Or

$y=-2/3$
Answer

$x=3/2$ and

$y=-2/3$

Solve for $x$ and $y$ :
$7(y+3)-2(x+2)=14$ ,
$4(y-2)+3(x-3)=2$ .

Simplify given equations

$7(y+3)-2(x+2)=14$
or

$7y-2x=-3$
and

$4(y-2)+3(x-3)=2$
or

$4y+3x=19$

new set of equations is :

$7y-2x=-3 ...(1)$

$4y+3x=19 ...(2)$

Let us use the elimination method to solve the given system of equation.
Multiply

$(1)$ by

$3$ and

$(2)$ by

$2$ . And add both the
equations.
Adding

$\dfrac { 21y-6x=-9\\ 8y+6x=38 }{ 29y\quad =29\Rightarrow y=\dfrac { 29 }{ 29 } =1 }$

Substitute the value of

$x$ in equation

$(1)$ , we have

$7(1)-2x=-3$

$x=5$

Answer:

$x=5$ and

$y=1$

Solve each of the following system of equations graphically and find the vertices and area of the triangle formed by these lines and the $x-$ axis:
$4x-3y+4=0,\ 4x+3y-20=0$

$4x-3y+4=0$ ....(1)

$4x+3y-20=0$ ......(2)

Isolate

$x$ from equation (1) and find the value of

$x$ and

$y$ .

$4x-3y+4=0$

$\Rightarrow x=\dfrac{3y-4}{4}$

$x$	$y$
$-1$	$0$
$2$	$4$
$-4$	$-4$

Similarly, isolate

$x$ from equation (2) and find the values of

$x$ and

$y$ .

$4x+3y-20=0$

$\Rightarrow x=\dfrac{20-3y}{4}$

$x$	$y$
$5$	$0$
$2$	$4$
$-1$	$8$

Graph:

On plotting

$(-1,0),\ (2,4)\text{ and } (-4,-4)$ on the

$xy-plane$ , we will get the graph of the linear equation

$4x-3y+4=0$ .

Similarly, on plotting

$(5,0),\ (2,4)\text{ and } (-1,8)$ on the

$xy-plane$ , we will get the graph of the linear equation

$4x+3y-20=0$ .

The obtained graph is shown above.
We can see from the graph that both the lines intersect each other at point

$(2, 4)$ and

$x-$ axis at

$A(-1, 0)$ and

$D(5, 0)$ .

Area of

$\triangle BAD$ :
Area of

$\triangle BAD=\dfrac{1}{2}\times base \times altitude$

$\Rightarrow \dfrac{1}{2} \times 6\times 4$

$\Rightarrow 12\ square\ units$ .

Hence, the vertices of the triangle formed by the given lines and

$x-axis$ are

$A(-1,0),\ B(2,4)\text { and }D(5,0)$ , and the area of the triangle is

$12\ square\ units$ .

1602609_1713806_ans_3d5130e3c2fa4c7e9d0280b5eed6b9c2.png

Solve each of the following system of equations graphically and find the vertices and area of the triangle formed by these lines and the $x-$ axis:
$x-y+1=0,\ 3x+2y-12=0$

$x-y+1=0$ ....(1)

$3x+2y-12=0$ ......(2)
Isolate

$x$ from equation (1) and find the value of

$x$ and

$y$ .

$x-y+1=0$
or

$x=y-1$

$x$	$y$
$-1$	$0$
$0$	$1$
$1$	$2$

Similarly, isolate

$x$ from equation (2) and find the values of

$x$ and

$y$ .

$3x+2y-12=0$
or

$x=(12-2y)/3$

$x$	$y$
$4$	$0$
$2$	$3$
$0$	$6$

Graph:
Both the lines intersect each other at point

$E(2, 3)$ and

$x-$ axis at

$A(-1, 0)$ and

$D(4, 0)$
To find the area of triangle

$BAD$ :
Area of

$\triangle EAD=1/2(base \times altitude)$

$=1/2 \times 5\times 3$

$=7.5\ sq.units$ .

1602642_1713809_ans_3975fda31db94d39b053aa3087fcd5c5.png

Solve for $x$ and $y$ :
$\dfrac5x+6y=13\\\dfrac3x+4y=7\quad(x\neq 0)$

Given

$\dfrac5x+6y=13\quad...(1)\\\dfrac3x+4y=7\quad...(2)$

Let us use elimination method to solve the given system of equation.

Multiply equation

$(1)$ by

$2$ and equation

$(2)$ by

$3$ , we get:

$\dfrac { 10 }{ x } +12y=26\quad...(3)\\ \dfrac { 9 }{ x } +12y=21\quad...(4)$

Now subtracting equation

$(4)$ from

$(3)$

$\left(\dfrac { 10 }{ x } +12y\right)-\left(\dfrac { 9 }{ x } +12y\right)=26-21$

$\Rightarrow\dfrac1x=5$

$\Rightarrow x=\dfrac15$

Substituting the value of

$x$ in equation

$(1)$ , we get:

$5\left(\dfrac15\right)+6y=13$

$\Rightarrow y=-2$

Hence,

$x=\dfrac15$ and

$y=-2$ .

Solve for $x$ and $y$ :
$0.3x+0.5y=0.5$
$0.5x+0.7y=0.74$ .

$0.3x+0.5y=0.5$

$0.5x+0.7y=0.74$ .

Multiply both the equations by

$10$ , we get

$3x+5y=5 ...(1)$

$5x+7y=7.4 ...(2)$

Multiply

$(1)$ by

$7$ and

$(2)$ by

$5$ . And subtract both the equations.

$\dfrac { 21x+35y=35\\ 25x+35y=37\\ -\quad -\quad - }{ -4x=\quad -2 }$

$x=\dfrac {-2}{-4}=\dfrac {1}{2}=0.5$

Substitute the value of

$x$ in equation

$(1)$ , we have

$3(0.5)+5y=5$

$y=0.7$

Answer :

$x=0.5$ and

$y=0.7$

Solve for $x$ and $y$ :
$2x+3y+1=0$ ,
$(7-4x)/3=y$

Let

$2x+3y+1=0 ...(1)$

$\dfrac{(7-4x)}3=y ...(2)$

Put value of

$y$ in

$(1)$ , we get

$2x+3\left (\dfrac{7-4x}3\right )+1=0$

$2x+7-4x+1=0$

$x=4$

from equation

$(2)$ :

$y=(7-4(4))/3$

$y=-3$

Answer :

$x=4$ and

$y=-3$

Solve for $x$ and $y$ :
$2x-\dfrac3y=9$
$3x+\dfrac7y=2\quad(y\neq 0)$

Given

$2x-\dfrac3y=9\quad...(1)\\3x+\dfrac7y=2\quad...(2)$

Let us use the elimination method to solve the given system of equation.

Multiply

$(1)$ by

$3$ and

$(2)$ by

$2$ ,We get

$6x-\dfrac9y=27\quad...(3)\\6x+\dfrac{14}y=4\quad...(4)$

Now, subtracting equation

$(3)$ from

$(4)$

$\left(6x+\dfrac{14}y\right)-\left(6x-\dfrac9y\right)=4-27$

$\Rightarrow\dfrac{14}y+\dfrac9y=-23$

$\Rightarrow\dfrac{23}y=-23$

$\Rightarrow y=-1$

Substitute the value of

$y$ in equation

$(1)$ , we have

$2x-\dfrac3{(-1)}=9$

$\Rightarrow 2x+3=9$

$\Rightarrow 2x=6$

$\Rightarrow x=3$

Hence

$x=3, y=-1$

Show graphically that each of the following given system of equation has infinitely many solutions:
$2x+3y=6,\ 4x+6y=12$

$2x+3y=6$ ....(1)

$4x+6y=12$ ......(2)
Isolate

$x$ from equation (1) and find the value of

$x$ and

$y$ .

$2x+3y=6$
or

$x=(6-3y)/2$

$x$	$y$
$3$	$0$
$0$	$2$
$-3$	$4$

Similarly, isolate

$x$ from equation (2) and find the values of

$x$ and

$y$ .

$4x+6y=12$
or

$x=(12-6y)/4$

$x$	$y$
$3$	$0$
$3/2$	$1$
$-3/2$	$3$

Graph:
From the graph, we can see that, all the points lie on the same straight line.
Which shows that, system has infinite many solutions.

1602739_1713854_ans_ab5e8b8a00a94e8eb013a2d25140b0a1.png

Solve each of the following system of equations graphically and find the vertices and area of the triangle formed by these lines and the $y$ -axis:
$4x-y-4=0,\ 3x+2y-14=0$

Given equations are

$4x-y-4=0\quad...(1)$

$3x+2y-14=0\quad...(2)$

Write

$y$ in terms of

$x$ for equation

$(1)$ .

$4x-y-4=0$

$\Rightarrow y=(4x-4)$

Substitute different values of

$x$ in the above equation to get corresponding values of

$y$

For

$x=1, \quad y=0$

For

$x=0, \quad y=-4$

For

$x=2, \quad y=4$

Now plot the points

$A(1,0),\ B(0,-4)$ and

$C(2,4)$ in the graph paper and join

$A,\ B$ and

$C$ to get the graph of

$4x-y-4=0$

Similarly, Write

$y$ in terms of

$x$ for equation

$(2)$ .

$3x+2y-14=0$

$\Rightarrow y=\dfrac{(14-3x)}2$

Substitute different values of

$x$ in the above equation to get corresponding values of

$y$

For

$x=0, \quad y=7$

For

$x=2, \quad y=4$

For

$x=4, \quad y=1$

Now plot the points

$D(0,7),\ E(2,4)$ and

$F(4,1)$ in the graph paper and join

$D,\ E$ and

$F$ to get the graph of

$3x+2y-14=0$

From the graph:

Both the lines intersect each other at point

$C(2, 4)$ and

$y$ -axis at

$B(0, -4)$ and

$D(0, 7)$ respectively

$\therefore$ Area of

$\triangle BCD=\dfrac12(\text{base} \times \text{altitude})$

$=\dfrac12 \times 11\times 2\ \text{sq.units}$

$=11\ \text{sq.units}$ .

1648003_1713831_ans_ed20cc0d9a774c2f8a5a4b2b00b91430.png

Solve each of the following system of equations graphically and find the vertices and area of the triangle formed by these lines and the $x-$ axis:
$x-2y+2=0,\ 2x+y-6=0$

Line 1:

$x-2y+2=0$ ....(1)
Line 2:

$2x+y-6=0$ ......(2)
Isolate

$x$ from equation (1) and find the value of

$x$ and

$y$ .

$x-2y+2=0$
or

$x=2y-2$

At $x=-2,0,2$	$\Rightarrow y=0,1,2$ respectively

Similarly, isolate

$y$ from equation (2) and find the values of

$x$ and

$y$ .

$2x+y-6=0$
or

$y=6-2x$

At $x=0,2,3$	$\Rightarrow y=6,2,0$ respectively

By Plotting the Graph we found that:
Both the lines intersect each other at point

$C(2, 2)$ and

$y-$ axis at

$A(-2, 0)$ and

$F(3, 0)$
Now, To find the area of triangle

$CAF$ :
Area of

$\triangle CAF=\dfrac{1}{2}(base \times altitude)$

$=\dfrac{1}{2} \times (3-(-2))\times (2-0)$

$=5\ sq.units$ .
1602698_1713813_ans_de9e95374b7243cc86a606ff365603af.png

Find two numbers such that the sum of twice the first and thrice the second is $92$ , and four times the first exceeds seven times the second by $2$ .

Let us consider, first number

$=x$ and

Second number

$=y$

As per statement,

$2x+3y=92 ... (1)$

$4x-7y=2 ...(2)$

Using elimination method

Multiply

$(1)$ by

$2$ and

$(2)$ by

$1$ , we get

$4x+6y=184 ...(3)$

$4x-7y=2 ...(4)$

Subtracting

$(3)$ from

$(4)$

$13y=182$

$y=14$

From

$(1), 2x+3y=92$

$2x+3\times 14=92$

$2x+42=92$

$2x=92-42=50$

$x=25$

First number

$=25$

Second number

$=14$

Find two numbers such that the sum of thrice the first and the second is $142$ , and four times the first exceeds the second by $138$ .

Let us consider,

First number

$=x$ and

Second number

$=y$

As per the statement,

$3x+y=142 ...(1)$

$4x-y=138 ...(2)$

Using elimination method

Add

$(1)$ and

$(2)$ , we get

$7x=280$

$x=40$

From

$(1):3\times 40+y=142$

$120+y=142$

$y=142-120=22$

Answer

First number

$40$

Second number

$=22$

A lady has only $25$ paisa and $50$ paisa coins in her purse. If she has $50$ coins in all totalling $Rs.\ 19.50$ , how many coins of each kind does she have?

Let us consider,
Number of

$25-$ paisa coins

$=x$ and
Number

$50-$ paisa coins

$=y$
Total number of coins

$=50$
and total amount

$=Rs\ 19.50$ or

$1950$ paisa

$x+y=50 ... (1)$

$25x+50y=1950$

$x+2y=78 ... (2)$
Subtracting

$(1)$ from

$(2)$

$y=28$
Ans,

$x=50-y=50-28=22$
Number of

$25-$ paisa coins

$=22$
and

$50-$ paisa coins

$=28$

The sum of a two digit number and the number obtained by receiving the order of its digits is $121$ , and the two digits differ by $3$ . Find the number.

Let us consider, one's digit of the two digit number

$=x$ and Ten's digit

$=y$

The number is

$x+10y$

By reversing the digits,

One's digit

$=y$ and

ten's digit

$=x$

The number is

$y+10x$

Now, As per the statement,

$x+10y+y+10x=121$

$11x+11y=121$

$x+y=11$ .....(1)

And,

$x-y=3$ .....(2)

On adding, we get

$2x=14$ or

$x=7$

On subtracting (2) from (1),

$2y=8$ or

$y=4$

Answer:

Number

$=7+10(4)=47$

$4+10(7)=4+70=74$

Show that the following systems of equation has a unique solution and solve it:
$x/2+y/2=9, x-2y=2$

$2x+3y=18 ....(1)$

$x-2y=2 ...(2)$

$a_1=2 , b_1=3, c_1=18$

$a_2=1 , b_2=-2, c_2=2$

$\dfrac {a_1}{a_2}=\dfrac {2}{1}=2, \dfrac {b_1}{b_2}=\dfrac {3}{-2}$

$\dfrac {a_1}{a_2}\neq \dfrac {b_1}{c_2}$

This system has unique solution.

Solve the equations: From

$(2), x=2+2y$

Substituting the value of

$x$ in

$(1)$ .

$2(2+2y)+3y=18$

$\Rightarrow 4+4y+3y=18$

$\Rightarrow 7y=18-4=14$

$\Rightarrow y=2$

and

$x=2+2\times 2=2+4=6$

Answer:

$x=6$ and

$y=2$ .

The sum of two numbers is $80$ . The larger number exceeds four times the smaller one by $5$ . Find the numbers.

Let

$x$ be the first number and

$y$ be the second number

As per statement ,

$x+y=80$ ......(1)

and

$x-4y=5$ .......(2)

On subtracting both the equations, we get

$y=15$

From

$(1): x+15=80$

$x=80 -15=65$

Answer:

Required numbers are

$15$ and

$65$ .

Show graphically that each of the following given system of equation has infinitely many solutions:
$2x+y=6,\ 6x+3y=18$

$2x+y=6$ ....(1)

$6x+3y=18$ ......(2)
Isolate

$y$ from equation (1) and find the value of

$x$ and

$y$ .

$2x+y=6$
or

$y=6-2x$

$x$	$y$
$0$	$6$
$2$	$2$
$4$	$-2$

Similarly, isolate

$x$ from equation (2) and find the values of

$x$ and

$y$ .

$6x+3y=18$

$x=(18-3y)/6$

$x$	$y$
$3$	$0$
$1$	$4$
$5$	$-4$

Graph:
Both the lines coincide each other.
This system has infinitely many solutions.

1602791_1713858_ans_9ffd49bd9ff74f5e86ef0cd5429a933b.png

Show graphically that the following given system of the equation has infinitely many solutions:
$3x-y=5,\ 6x-2y=10$

$3x-y=5$ ....(1)

$6x-2y=10$ ......(2)
Isolate

$x$ from equation (1) and find the value of

$x$ and

$y$ .

$3x-y=5$
or

$y=3x-5$

At,

$x$	$y$
$0$	$-5$
$1$	$-2$
$3$	$4$

Similarly, isolate

$x$ from equation (2) and find the values of

$x$ and

$y$ .

$6x-2y=10$

$y=(6x-10)/2$

At,

$x$	$y$
$2$	$1$
$4$	$7$
$3$	$4$

Using the above points and plotting the Graph we get,
Both the lines coincide with each other that means that system has infinite many solutions.

1602755_1713856_ans_658878bc6acf4805bf301e5d3bb054dc.png

$5$ chairs and $4$ tables together cost $Rs.\ 5600$ , while $4$ chairs and $3$ tables together cost $Rs.\ 4340$ . Find the cost of a chair and that of a table.

Let us consider,

Cost of one chair

$=Rs\ x$ and

Cost of one table

$=Rs\ y$

As per the statement,

$5x+4y=Rs\ 5600 ... (1)$

$4x+3y=Rs\ 4340 .... (2)$

Using elimination method:

Multiply

$(1)$ by

$(3)$ and

$(2)$ by

$4$ . And subtract both the equations.

$15x+12y=16800$

$16x+12y=17360$

$-$

_________________

$-x$

$=-560$

So,

$x=560$

From

$(1)$

$5\times 560+4y=5600$

$2800+4y=5600$

$4y=5600-2800$

$4y=2800$

$y=700$

Answer

Cost of one chair

$=Rs\ 560$

and cost of one table

$=Rs\ 700$

Show graphically that each of the following given system of equation has infinitely many solutions:
$x-2y=5,\ 3x-6y=15$

Let,

$x-2y=5$ ....

$(1)$

$3x-6y=12$ ......

$(2)$
Isolate

$x$ from equation

$(1)$ and find the value of

$x$ and

$y$ .

$x-2y=5$
or

$x=5+2y$

$x$	$y$
$5$	$0$
$3$	$-1$
$1$	$-2$

Similarly, isolate

$x$ from equation (2) and find the values of

$x$ and

$y$ .

$3x-6y=15$
or

$x=(15+6y)/3$

$x$	$y$
$7$	$1$
$-1$	$-3$
$-3$	$-4$

Graph:
All the points lie on the same line.
Which shows that, lines coincide each other. Hence, the system has infinite many solutions.

1602930_1713863_ans_7e0fd1dda0cd4fbea1b86ffd4c40bdf1.png

$\dfrac{4}{3}x+2y=8\ ;$
$2x+3y=12$

We have,

$$\dfrac{4}{2}x+2y=8\qquad...(i)\\2x-3y=12\qquad...(ii)$$

Here,

$a_1=\dfrac{4}{3},b_1=2,c_1=8$ and

$a_2=2,b_2=3,c_2=12$

Thus,

$\dfrac{a_1}{a_2}=\dfrac{4}{3\times2}=\dfrac{2}{3},\quad \dfrac{b_1}{b_2}=\dfrac{2}{3},\quad \dfrac{c_1}{c_2} = \dfrac{8}{12}=\dfrac{2}{3}$

$\therefore \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2},$ so equations (i) and (ii) represent coincident lines.

Hence, the pair of linear equations is consistent with infinitely many solutions.

From the pair of linear equations in this problem, and find its solution graphically:
10 students of class X took part in a Mathematics quiz If the girls is 4 more than the number of boys, find the number of boy and girls who took part in the quiz.

Let

$x$ be the number of girls and

$y$ be the number of boys.

According to the questions, we have

$\quad x=y+4$

$\Rightarrow x-y=4\qquad...(i)$

Substituting different values of

$x$ in equation

$(i)$ to get corresponding values of

$y$ , we get the following table :

$x$	$0$	$4$	$7$
$y$	$-4$	$0$	$3$

Plot the points

$(0,-4),(4,0),(7,3)$ in the graph paper and join them to get the line represented by

$x-y=4$

Again, the total number of students

$=10$

$\Rightarrow x+y=10\qquad...(ii)$

Substituting different values of

$x$ in equation

$(ii)$ to get corresponding values of

$y$ , we get the following table :

$x$	$0$	$10$	$7$
$y$	$10$	$0$	$3$

Plot the points

$(0,10),(10,0),(7,3)$ in the graph paper and join them to get the line represented by

$x+y=10$

From the graph,

The two lines intersect at the point

$(7,3)\ \ \ \Rightarrow x=7\ ,\ y=3$

Hence, the number of girls

$=7$

And the number of boys

$=3$

1703470_1786731_ans_c1ae6a5bd40440c39f136a2637ee6b21.png

Solve the following pairs of equations by reducing them to a pair of linear equations:
(i) $\dfrac{7x-2y}{xy}=5\ \ ;\ \ \dfrac{8x+7y}{xy}=15$
(ii) $\dfrac{1}{3x+y}+\dfrac{1}{3x-y}=\dfrac{3}{4}\ \ ;\ \ \dfrac{1}{2(3x+y)}-\dfrac{1}{2(3x-y)}=\dfrac{-1}{8}$

(i) We have

$\dfrac{7x-2y}{xy}=5 \Rightarrow \dfrac{7x}{xy}-\dfrac{2y}{xy}=5\\ \Rightarrow \dfrac{7}{y}-\dfrac{2}{x}=5$

$\text{And, }\dfrac{8x+7y}{xy}=15 \Rightarrow \dfrac{8x}{xy}+\dfrac{7y}{xy}=15\\ \Rightarrow \dfrac{8}{y}+\dfrac{7}{x}=15$

Let

$\dfrac{1}{y}=u$ and

$\dfrac{1}{x}=v$

Then,

$7u-2v=5\quad...(i)$

$8u+7v=15\quad...(ii)$

Multiplying equation

$(i)$ by

$7$ and

$(ii)$ by

$2$ and adding, we have

$49u-14v=35$

$16u+14v=30$

$\overline{65u\ +0\ \ =65}$

$\therefore u=\dfrac{65}{65}=1$

Putting the value of u in equation

$(i)$ , we get

$7\times-2u=5 \Rightarrow -2v=5-7=-2$

$\therefore -2v=-2\Rightarrow v=\dfrac{-2}{-2}=1$

Now,

$u=1 \Rightarrow \dfrac{1}{y}=1 \Rightarrow y=1$

And,

$v=1 \Rightarrow \dfrac{1}{x}=1 \Rightarrow x=1$

Hence, the solution of given system of equations is

$x=1\ ,\ y=1$ .

(ii) We have,

$\dfrac{1}{3x+y}+\dfrac{1}{3x-y}=\dfrac{3}{4}$ and

$\dfrac{1}{2(3x+y)}-\dfrac{1}{2(3x-y)}=-\dfrac{1}{8}$

Let,

$\dfrac{1}{3x+y}=u$ and

$\dfrac{1}{3x-y}=v$

Then,

$u+v=\dfrac{3}{4}\qquad...(i)$ and

$\dfrac{u}{2}-\dfrac{u}{2}=-\dfrac{1}{8} \\\Rightarrow \dfrac{u-v}{2}=-\dfrac{1}{8}\\\Rightarrow u-v=-\dfrac{2}{8}=-\dfrac{1}{4}\\\Rightarrow u-v=-\dfrac{1}{4}\qquad...(ii)$

Adding

$(i)$ and

$(ii)$ , we get

$u+v=\dfrac{3}{4}$

$u-v=-\dfrac{1}{4}$

$\overline{2u=\dfrac{3}{4}-\dfrac{1}{4}}$

$\Rightarrow 2u=\dfrac{3-1}{4}=\dfrac{2}{4}$

$\Rightarrow u=\dfrac{2}{4\times2}=\dfrac{1}{4} \qquad \\\ \therefore u =\dfrac{1}{4}$

Now putting the value of

$u$ in equation

$(i)$ , we get

$\dfrac{1}{4}+v=\dfrac{3}{4} \Rightarrow v=\dfrac{3}{4}-\dfrac{1}{4}=\dfrac{3-1}{4}=\dfrac{2}{4}=\dfrac{1}{2} \Rightarrow v=\dfrac{1}{2}$

Now,

$v=\dfrac{1}{4} \Rightarrow \dfrac{1}{3x+y}=\dfrac{1}{4} \Rightarrow 3x+y=4\quad...(iii)$

And,

$v=\dfrac{1}{2} \Rightarrow \dfrac{1}{3x+y}=\dfrac{1}{2}\Rightarrow 3x-y=2\quad...(iv)$

Now, adding

$(iii)$ and

$(iv)$ , we get

$3x+y=2$

$3x-y=2$

$\overline{\ \ \ \ \ \ 6x \ \ \ =6\ \ \ }$

$\Rightarrow x=\dfrac{6}{6}=1$

Putting the value of

$x$ in equation

$(iii)$ , we get

$3\times 1+y=4$

$\Rightarrow y=4-3=1$

Hence, the solution of given system of equation is

$x=1\ ,\ y=1$

The sum of two numbers is $60$ and their difference is $30$ .
The number are and _.

We have,

$x=15 \ \ \textbf{[From (C)]}$
Therefore, other number

$=60-x$
Hence,

$(60-x)=60-15$

$=45$

$\textbf{Hence, the numbers are 15 and 45.}$

$5x-4y+8=0\\7x-6y-9=0$

We have ,

$5x-4y+8=0\quad...(i)\\7x+6y-9=0\quad...(ii)$

Here,

$a_1=5, b_1=-4, c_1=8$ and

$a_2=7, b_2=6, c_2=-9$

$\Rightarrow \dfrac{a_1}{a_2}=\dfrac{5}{7}$ and

$\dfrac{b_1}{b_2}=-\dfrac{4}{6}=-\dfrac{2}{3}$

Since

$\dfrac{a_1}{a_2}\neq \dfrac{b_1}{b_2}$ , the given pair is consistent with a unique solution.

Hence, the given pair represents two lines intersecting at a point.

$\dfrac{3}{2}x+\dfrac{5}{3}y=7\ ;$
$9x-10y=14$

We have ,

$\dfrac{3}{2}x+\dfrac{5}{3}y=7\qquad...(i)$

$9x-10y=14\qquad...(ii)$

Here

$a_1=\dfrac{3}{2},b_1=\dfrac{3}{5},c_1=7$

$a_2=9,b_2=-10,c_2=14$

Thus,

$\dfrac{a_1}{a_2}=\dfrac{3}{2\times9}=\dfrac{1}{6}\ \ ,\ \ \dfrac{b_1}{b_2}=\dfrac{5}{3\times(-10)}=-\dfrac{1}{6}$

$\therefore \dfrac{a_1}{a_2}\neq \dfrac{b_1}{b_2}.$

So, it has a unique solution and it is consistent.

Is the following pair of linear equations consistent? Justify your answer.
$2ax+by=a,\quad 4ax+2by-2a=0;\quad a,b \neq0$

We know that,

For a pair of linear equations to be consistent,

$\ \ \dfrac{a_1}{a_2}\neq\dfrac{b_1}{b_2}$ [Unique solution]

or,

$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$ [Infinite solutions]

Here,

$\dfrac{a_1}{a_2}=\dfrac{2a}{4a}=\dfrac{1}{2}, \quad \dfrac{b_1}{b_2}=\dfrac{b}{2b}=\dfrac{1}{2},\quad \dfrac{c_1}{c_2}=\dfrac{-a}{-2a}=\dfrac{1}{2}$

$\therefore \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$

Hence, The given system of equations is consistent and has infinite solutions.

Solve the following pairs of linear equation by the elimination method and the substitution method:
(i) $3x-5y-4=0$ and $9x=2y+7$
(ii) $\dfrac{x}{2}+\dfrac{2y}{3}=-1$ and $x-\dfrac{y}{3}=3$

$(i)$ We have,

$3x-5y-4=0$

$\Rightarrow 3x-5y=4 \qquad...(i)$

Again

$9x=2y+7$

$\Rightarrow 9x-2y=7\qquad...(ii)$

$\text{By Elimination Method:}$

Multiplying equation

$(i)$ by 3, we get

$9x-15y=12\qquad...(iii)$

Subtracting

$(ii)$ from

$(iii)$ , we get

$9x-15y=12$

$9x-2y=7$

$\overline{\ \ \ -13y=5\ \ \ \ \ }$

$\Rightarrow y=-\dfrac{5}{13}$

Putting the value of equation

$(ii)$ , we get

$9x-2\left(-\dfrac{5}{13}\right)=7\\ \Rightarrow 9x+\dfrac{10}{13}=7\\\Rightarrow 9x=7-\dfrac{10}{13}\\\Rightarrow 9x=\dfrac{91-10}{13}\\\Rightarrow 9x=\dfrac{81}{13}\\\Rightarrow x=\dfrac{9}{13}$

Hence, the required solution is

$x=\dfrac{9}{13},y=-\dfrac{5}{13}$

$\text{By Substitution Method:}$

Expressing

$x$ in terms of

$y$ from equation

$(i)$ , we have

$x=\dfrac{4+5y}{3}$

Substituting the value of

$x$ in equation

$(ii)$ , we get

$9\times\left( \dfrac{4+5y}{3}\right)-2y=7\\ \Rightarrow 3\times(4+5y)-2y=7\\ \Rightarrow 12+15y-2y=7 \\\Rightarrow 13y=7-12\\\therefore y=-\dfrac{5}{13}$

Putting the value of

$y$ in equation

$(i)$ , we have

$3x-5\times\left(-\dfrac{5}{13}\right)=4 \\\Rightarrow 3x+\dfrac{25}{13}=4\\\Rightarrow 3x=4-\dfrac{25}{13}\\ \Rightarrow 3x=\dfrac{27}{13}\\\therefore x=\dfrac{9}{13}$

Hence, the required solution is

$x=\dfrac{9}{13}\ ,\ y=-\dfrac{5}{13}.$

$(ii)$ We have,

$\dfrac{x}{2}+\dfrac{2y}{2}=-1$

$\Rightarrow \dfrac{3x+4y}{6}=-1$

$\therefore 3x+4y=-6...(i)$

And

$x-\dfrac{y}{2}=3 \Rightarrow \dfrac{3x-y}{3}=3$

$\therefore 3x-y=9\qquad...(ii)$

$\text{By Elimination Method:}$

Subtracting

$(ii)$ from

$(i)$ , we get

$5y=-15\ \Rightarrow y=-\dfrac{15}{5}=-3$

Putting the value of

$y$ in equation

$(i)$ , we get

$3x+r\times(-3)=-6\\ \Rightarrow 3x-12=-6\\\Rightarrow 3x=-6+12\Rightarrow 3x=6$

Hence, Solution is

$x=2\ ,\ y=-3$

$\text{By Substitution Method:}$

Expressing

$x$ in terms of

$y$ from equation

$(i)$ , we have

$x=\dfrac{-6-4y}{3}$

Substituting the value of

$x$ in equation

$(ii)$ from equation

$(i)$ , we get

$3\times\left(\dfrac{-6-4y}{3}\right)-y=9 \\\Rightarrow -6-4y-y=9 \\\Rightarrow -6-5y=9 \\\Rightarrow -5y=9+6=15\\\therefore y=\dfrac{15}{-5}=-3$

Putting the value of

$y$ in equation

$(i)$ , we get

$3x+\times(-3)=-6\\ \Rightarrow 3x-12=-6 \\\Rightarrow 3x=12-6=6 \\\therefore x=\dfrac{6}{3}=2$

Hence, the required solution is

$x=2,y=-3$

Find whether the following pair of linear equations has a unique solution. If yes, find the solution?
$7x-4y=49$ and $5x-6y=57$

We have,

$7x-4y=49\qquad...(i)$ and

$5x-6y=57...(ii)$

Here

$a_1=7,b_1=-4,c_1=49$ and

$a_2=5,b_2=-6,c_2=57$

So,

$\dfrac{a_1}{a_2}=\dfrac{7}{5}\ ,\ \dfrac{b_1}{b_2}=\dfrac{-4}{-6}=\dfrac{2}{3}$

Since,

$\dfrac{a_1}{a_2}\neq \dfrac{b_1}{b_2}$

The system has a unique solution.

Now, Multiply equation

$(i)$ by

$5$ and equation

$(ii)$ by

$7$ and subtract,

$35x-20y=245$

$35x-42y=399$

$\overline{\quad 22y=-154 \quad}\\ \Rightarrow \quad y=-7$

Put

$y=-7$ in equation

$(ii)$

$5x-6(-7)=57\\ \Rightarrow 5x=57-42 \quad \\\Rightarrow x=3$

Hence

$x=3$ and

$y=-7$

Show graphically that the given system of equations $2x+4y=10$ and $3x+6y=12$ has no solution.

Given equations are,

$2x+4y=10$ and

$3x+6y=12$

Now,

$2x+4y=20$

$\Rightarrow 4x=10-2x \\\Rightarrow y=\dfrac{5-x}{2}$

Thus, we have the following table

$x$	$3$	$1$	$5$
$y$	$1$	$2$	$0$

Plot the points

$A(1,2), B(3,1)$ and

$C(5.0)$ on the graph paper, Join

$A,B$ and

$C$ and extent it on both sides to obtain the equation

$2x+4y=10$ .

Again,

$3x+6y=12$

$\Rightarrow 6y=12-3x \\ \Rightarrow y=\dfrac{5-x}{2}$

Thus, we have the following table

$x$	$2$	$0$	$4$
$y$	$1$	$2$	$0$

Plot the points

$D(2,1), E(0,2)$ and

$F(4,0)$ on the same graph paper. Join

$D,E$ and

$F$ and extend on both sides to obtain the graph of the equation

$3x+6y=12$

We find that the lines represented by equation

$2x+4y=10$ and

$3x+6y=12$ are parallel.

So the two lines have no common point.

Hence, the given system of equations has no solution.

1701099_1786758_ans_e427c28bdd39499bae32f4d94a93e85a.png

Solve the following pairs of linear equations :

$4x +\dfrac {6}{y} =15, 6x - \dfrac {8}{y} = 14, y \neq 0$

. $4x + \dfrac {6}{y} = 15 .....(i) \times 6$ or $3$

$6x - \dfrac {8}{y} = 14 ...(ii) \times 4$ or $2$

(as $y$ in denominators and symmetric so no need to remove denominator and $6$ and $4$ are divisible by $2$ so we can multiply $(i) , (ii)$ by $3$ and $2$ respectively.)

$12 x + \dfrac {18}{y} = 45 ....(iii)$

$12x - \dfrac {16}{y} = 28......(iv)$

$- \quad + \quad -$

___________________

$\dfrac {18}{y} + \dfrac {16}{y} = 17$ [ By subtracting $(iv)$ from $(ii)$ ]

$\dfrac { 18 +16}{y} = \dfrac {17}{1 }$

$\Rightarrow 17y = 34$

$\Rightarrow y = \dfrac {34}{ +17} \ Rightarrow y = 2$

Now $12x + \dfrac {18}{y}= 45$ [ from (iii) ]

$\Rightarrow 12x + \dfrac { 18}{2} = 45$

$\Rightarrow 12x = 45- 9$

For which values of $a$ and $b$ , will the following pair of linear equations has infinitely many solution $Rs$
$x +2y = 1$
$(a-b)x + (a+b)y = a+b -2$

For infinitely many solutions,

$\dfrac {a_1}{a_2} = \dfrac {b_1}{b_2} = \dfrac {c_1}{c_2}$

$\Rightarrow \dfrac { 1 }{ \begin{matrix} (a-b) \\ I \end{matrix} } =\dfrac { 2 }{ \begin{matrix} (a\quad +b) \\ II \end{matrix} } \neq \dfrac { 1 }{ \begin{matrix} (a+b-2) \\ III \end{matrix} }$

From ratio

$I$ and

$II$

$2a -2b = a+b$

$\Rightarrow a -3b = 0....(i)$

From ratio

$II$ and

$III$

$2a +2b - 4 = a+b$

$\Rightarrow a +b = 4 .....(ii)$

Now solving

$(i)$ and

$(ii)$ we have

$a -3b = 0 ......(i)$

$a + b = 4......(ii)$ [Subtracting (ii) from (i) ]

$- \quad - \quad -$
___________________

$-4b = -4$

$\Rightarrow b =1$
and

$a = 4 - b$

$\Rightarrow a = 4-1 [ from (ii) ]$

$\Rightarrow a = 3$

IF $2x +y = 23$ and $4x-y=19$ then find the values of $5y -2x$ and $(y/x)-2$

Given equations are $2x +y = 23 ...(i)$

$4x - y = 19 ....(ii)$

Adding equations are $(i)$ and $(ii)$ we get

$6x = 42$

$\Rightarrow x = 7$

Now $2(7) + y = 23$ [From}

$\Rightarrow y = 23-14$

$\Rightarrow y = 9$ $

So $5y =2x = 5(9) - 2(7)= 45 -14 = 31$

and $\dfrac {Y}{x} -2 = \dfrac {9}{7} -2 = \dfrac { 9-14}{7} = \dfrac {-5}{7}$

Hence the values of $(5y - 2x)$ and $( \dfrac {y}{x} -2 )$ are $31$ and $\dfrac {-5}{7}$ respectively.

Solve the following pairs of linear equations :

$\dfrac {1}{2x} -\dfrac {1}{y} = -1, \dfrac {1}{x} + \dfrac {1}{2y} = 8 , x,y \neq 0$

Given equation are

$\dfrac {1}{2x} - \dfrac {1}{y} = -1 ...(i) \times \dfrac {1}{2}$

$\dfrac {1}{x} + \dfrac {1}{2y} = 8 ...(ii) \times , 1 x,y \neq 0$

[

$x, y$ both are denominator and symmetric no need to convert into linear equation hence can be eliminated directly]

Multiplying eqn (i) by the coefficient of

$\dfrac {1}{y}$ in (ii) and vice versa m we have

$\dfrac {1}{4x} - \dfrac {1}{2y} = - \dfrac {1}{2} ...(iii)$

$\dfrac {1}{x} + \dfrac {1}{2y} = 8 ....(iv)$

_________________________

$\dfrac {1}{4x} + \dfrac {1}{x} = 8 -\dfrac {1}{2}$ [ adding eqn

$(iii)$ and

$(iv)$ ]

$\Rightarrow \dfrac {1+4}{4x} = \dfrac {16 -1}{2}$

$\Rightarrow {5}{4x} = \dfrac {15}{2}$

$\Rightarrow 15 \times 4x = 5 \times 2$

$\Rightarrow x = \dfrac { 5 \times 2 }{ 15 \times 4 } \Rightarrow x = \dfrac {1}{6}$

Now ,

$\dfrac { 1}{x} + \dfrac {1}{2y} = 8$ [ from (iv) ]

$\Rightarrow \dfrac {1}{ ( \dfrac {1}{6} ) } + \dfrac {1}{2y} = 8 [ \because x = \dfrac {1}{6} ]$

$\Rightarrow 6 + \dfrac {1}{2y} = 8 \Rightarrow \dfrac {1}{2y} = 8-6$

$\Rightarrow \dfrac {1}{2y} = \dfrac {2}{1} \Rightarrow 4y = 1 \Rightarrow = y = \dfrac {1}{4}$

So,

$x = \dfrac {1}{6}$ and

$y = \dfrac {1}{4}$ .

Solve the following pairs of linear equations :

$\frac {x}{3} + \dfrac {y}{4} = 4 , \dfrac {5x}{6}-\dfrac {y}{8} = 4$

$\dfrac {x}{3} + \dfrac {y}{4} = 4 ...(i) \times LCM\quad 12$

$\dfrac { 5x}{6} - \dfrac {y}{8} = 4 ...(ii) \times LCM \quad 24$

$4x +3y = 48 ......(iii)$

$20x -3y = 96 ......(iv)$

$24x = 144$ [ adding equations $iii$ and $iv$ ]

$\Rightarrow x = \dfrac {144}{24} \Rightarrow x = 6$

Now , $4x + 3y = 48$ [ from (iii) ]

On putting the value of $x =6$ we have

$4( 6) +3y = 48$

$\Rightarrow 3y = 48 -24$

$\Rightarrow 3y = 24$

$\Rightarrow y = \dfrac {24}{3} \Rightarrow y = 8$

So the solution of the given equations is $x = 6$ and $y = 8$

Find the values of $x$ and $y$ in the following rectangle :

$\textbf{Step 1: Find x and y}$

${3x + y = 7\text{.....(i)}}$

${x + 3y = 13\text{... ..(ii)}}$

$\text{Multiplying (i) with 3}$

$9x + 3y = 21 \text{ ....... (iii)}$

$\text{Subtracting equation (iii) from (ii)}$

$8x=8$

$\text{Hence, we see}$

$x = 1$

$\textbf{Step 2: Finding out the other term using the known term}$

$x = 1$

$\text{Putting value of x in eq(i), we get}$

$\;3\left( 1 \right) + y = 7\;\left[ \textbf{equation i } \right]\;$

$\Rightarrow y = 7 - 3$ $\Rightarrow y = 4$

$\text{and }x = 1$

$\textbf{Thus, the required values of x and y are 1 and 4 respectively.}$

Solve the following pairs of linear equations :

$x + y = 3.3 , \dfrac {0.6}{ 3x -2y} = -1, 3x -2y = 0$

We have

$x +y = 3.3 ....(i)$

$\dfrac { 0.6}{3x -2y} = -1 ......(ii)$
On multiplying eqn

$(i)$ by

$20$ we get

$20 x + 20 y = 66$
From eqn (ii) we have

$-3x + 2y = 0.6$
On multiplying it by

$( -10 )$ we get

$30 x - 20y = -6 ....(iii)$
Now adding

$(iii)$ and

$(iv)$ we get

$30 x - 20y = -6 .....(i)$

$20 x +20y = 66......(ii)$
______________________

$50x \quad =60$

$\Rightarrow x = \dfrac {60}{50} \Rightarrow x =\dfrac {6}{5} \Rightarrow = 1.2$
Now

$10( 1.2) 10 y = 33$ [ from (i) ]

$\Rightarrow 12 +10y =33$

$\Rightarrow 10y = 33 -12$

$\Rightarrow y = \dfrac {21}{10} \Rightarrow y = 2.1$

$\therefore$ The solution of the given system of equation is

$x = 1.2$ and

$y = 2.1$

Two straight paths are represented by the equations $x-3y = 2$ and $-2x +6y = 5$ check whether the paths will cross each other or not.

Two straight paths are represented by the equations

$x -3y = 2$ and $-2x + 6y = 5$

For the paths to cross each other i.e. intersect each other , we must have $\dfrac {a_1}{a_2} \neq \dfrac {b_1}{b_2}$

Now, $\dfrac {a_1}{a_2} = \dfrac {1}{-2} = \dfrac {-1}{2}$ and $\dfrac {b_1}{b_2} = \dfrac {-3}{6} = \dfrac {-1}{2}$

$\therefore \dfrac {a_1}{a_2} = \dfrac {b_1}{b_2}$

Hence , the two straight paths do not cross each other

Find the values of $p$ in (i) to (iv) and $p \quad q$ in (v) for the following pair of equations :

$2x +3y = 7$ and $px -6y - 8 = 0$ if the pair of equations has infinitely many solutions.

given system of linear equations is

$2px + py = 28 -qy$

$i.e. 2px + ( p +q)y = 28.....(i)$

$2x +3y = 7 ....(ii)$

The system of equations will have infinitely many solutions if

$\dfrac {a_1}{a_2} = \dfrac {b_1}{b_2} = \dfrac {c_1}{c_2}$

$\Rightarrow \dfrac { 2p }{ \begin{matrix} 2 \\ I \end{matrix} } =\dfrac { p+q }{ \begin{matrix} 3 \\ II \end{matrix} } =\dfrac { 28 }{ \begin{matrix} 7 \\ III \end{matrix} }$

using ratios $I$ and $II$ we get

$\dfrac {2p}{2} = \dfrac { p +q}{3}$

$\Rightarrow 3p = p +q$

$\Rightarrow 2p -q = 0$

$\Rightarrow q = 2p ......(iii)$

using ration $I$ and $III$ we get,

$\dfrac { 2p}{2} = \dfrac {28}{7} \Rightarrow p = 4$

$\therefore q = 2p = 2 \times 4 = 8$ [ From (iii) ]

$\therefore q = 8$ and $p = 4$

$Now, \dfrac {2p}{2} = \dfrac {p+q}{3} = \dfrac {28}{7}$

$\Rightarrow \dfrac {P}{1} = \dfrac {p+q}{3} = \dfrac {4}{1}$

By substituting the values of $P$ and $q$ we have

$4 = \dfrac { 4+ 8}{3} = 4$

$\Rightarrow 4 = \dfrac {12}{3} = 4$

$\Rightarrow 4 = 4 = 4$

hence the given system of equation has infinitely many solution when $p = 4$ and $q = 8$

For Which value(s) of $\lambda$ do the pair of linear equations $\lambda x + y + = \lambda^2$ and $x + \lambda y = 1$ have

a unique solution $Rs$

For unique solutions,

$\dfrac {a_1}{a_2} \neq \dfrac {b_1}{b_2}$

$\therefore \dfrac { \lambda}{1} \neq \dfrac { 1}{ \lambda}$

$\Rightarrow \lambda^2 \neq 1$ or

$\lambda \neq 1 , -1$
So the unique solution all real value except

$\lambda = 1 , -1$

Two years ago salim was thrice as old is daughter and six years later he will be four years older than twice her age .how old are they now $rs$ .

$\textbf{Step 1: Formulating the linear equations in 2 variables from the conditions given}$

$\text{Let the current age of Salim be }x\text{ and the current age of his daughter be }y$

$\text{Two years ago Salim was thrice as old as his daughter}$

$\Rightarrow x-2=3(y-2)$

$\Rightarrow x-2=3y-6\;\;............(1)$

$\text{Six years later he will be four years older than twice her age}$

$\Rightarrow x+6=2(y+6)+4$

$\Rightarrow x+6=2y+16\;\;...........(2)$

$\textbf{Step 2: Solving both equations to find each of their current ages}$

$x-2=3y-6$

$\Rightarrow x=3y-4\;\;................(3)$

$\text{Substituting 3 in 2}$

$3y-4+6=2y+16$

$y=14\;\;........................(4)$

$\text{Substituting }y\text{ in 3}$

$\Rightarrow x=8$

$\textbf{Hence, Salim's current age is 38 while his daughter's current age is 14}$ .

By the graphical method, find the whether the following pair of equation are consistent or not . if consistent solve them.

$3x +y +4 = 0$ and $6x -2y +4 = 0$

Given equation are

3x+y+4=0...(i)

1656306_1795682_ans_b548f7c197244255acc9cd0c5735e77a.PNG

Solve the following pairs of linear equations :

$\frac {2xy}{x+y} = \frac {3}{2}, \frac {xy}{2x-y} = \frac {-3}{10} , x+y \neq 0 ,2x-y \neq 0$

we have $\dfrac {2xy}{ x+ y} = \dfrac {3}{2} ....(i)$

$\dfrac {xy}{2x -y } = \dfrac {-3}{ 10 } ....(ii)$

$[ ( x+y) \neq 0$ and $( 2x -y ) \neq 0) ]$

Inversing the eqn $(i)$ we get

$\dfrac {x+y}{2xy} = \dfrac {2}{3}$

$\Rightarrow \dfrac {x}{2xy} + \dfrac {y}{2xy} = \dfrac {2}{3}$

$\Rightarrow \dfrac {1}{2y} + \dfrac {1}{2x} = \dfrac {2}{3} ....(iii)$

Inversing the eqn (ii) we get

$\dfrac { 2x - y }{xy } = \dfrac {10}{-3}$

$\Rightarrow \dfrac {2x}{xy} - \dfrac {y}{xy } = \dfrac {-10}{3}$

$\Rightarrow \dfrac {2}{y} - \dfrac {1}{x} = \dfrac {-10}{3}$

$\Rightarrow \dfrac {2}{2y} - \dfrac {1}{2x} = \dfrac {-5}{3} ...(iv)$

now, $\dfrac {2}{2y}- \dfrac {1}{2x} = \dfrac {-5}{3}$ [ from (iv) ]

$\dfrac {1}{2y} - \dfrac {1}{2x} = \dfrac {2}{3}$ [ from (ii) ]

and ______________________

$\dfrac {2}{2y} + \dfrac {1}{2y} = - \dfrac {5}{3} + \dfrac {2}{3}$ [ adding $(iii)$ and $(iv)$ ]

$\Rightarrow \dfrac { 2+1}{2y} = \dfrac { -5 +2 }{3} \Rightarrow \dfrac {3}{2y} = \dfrac {-3}{3}$

$\Rightarrow \dfrac {3}{2y} = -1 \Rightarrow y = - \dfrac {3}{3}$

now

$\dfrac {2}{2y} - \dfrac {1}{2x} = \dfrac {-5}{3}$

$\Rightarrow \dfrac {2}{y} - \dfrac {1}{x} = \dfrac {-10}{3}$

$\Rightarrow \dfrac {2}{ ( \dfrac {-3}{2} ) } - \dfrac {1}{x} = \dfrac {-10}{3} [ \because y = \dfrac {-3}{2} ]$

$\Rightarrow \dfrac {4}{-3} - \dfrac {1}{x} = \dfrac {-10}{3}$

$\Rightarrow - \dfrac {1}{x} = \dfrac {-10}{3} + \dfrac {4}{3}$

$\Rightarrow \dfrac {-1}{x} = \dfrac {-6}{3}$

$\Rightarrow \dfrac {1}{x} = \dfrac {+2}{1} \Rightarrow 2x = 1 \Rightarrow x = \dfrac {1}{2}$

So , $x = \dfrac {1}{2}$ and $y = \dfrac {-3}{2}$ .

The age of the father is twice the sum of the ages of his two children .after $20$ years,his age will be equal to the sum of the ages of his children . find the age of the father.

Let the present age of father be

$x$ years
Also , let the sum of present ages of two children by

$y$ years
The age of fathers is

$( = )$ twice

$( \times 2)$ the sum of the sum of the ages of two children

$\Rightarrow x = 2 \times (y) \Rightarrow x = 2y$

$\Rightarrow x -2y = 0$
Age of father

$20$ year later =

$( x +20)$ years
increase in age of first children in

$20$ years =

$20$ years
Increase in age of second children in

$20$ years =

$20$ years

$\therefore$ Increase in the age of both children in

$20$ years =

$20 +20 = 40$ years

$\therefore$ Sum of ages of both children

$20$ years later

$= ( y +40 )$
Now , according to the question , we have
father will be

$(=)$ sum of ages of two children[Given ]

$\Rightarrow x +20 = y +40$

$\Rightarrow x -y = 20 ...(ii)$

$\Rightarrow 2y - y = 20 [ (Rs x = 2y) From (i) ]$

$\Rightarrow y = 20$ years
Now ,

$x = 2y$ [ From (i) ]

$\Rightarrow x = 2 \times 20$

$\Rightarrow x = 40$

$\therefore$ Age of father is

$40$ years.

Draw the graph of the pair of equations $2x +y = 4$ and $2x -y= 4$ write the vertices of the triangle formed by these lines and the y-axis .also find the area of this triangle.

$2x +y = 4 ...(i)$

$\Rightarrow y = 4 -2 x$

$x = 0 , y = 4 - 2( 0 ) = 4 -0 = 4$

$x = 1 , y = 4-2 (1) = 4 -2 = 2$

$x = 2, y = 4 -2(2) = 4 -4 = 0$

$x$

$0$

$1$

$2$

$3$

$y$

$4$

$2$

$0$

$-2$

$(i)$

$A$

$B$

$C$

$-2$

$2x -y= 4 ...(ii)$

$\Rightarrow y = 2x - 4$

$x = 0 , y = 2 (0 ) -4 = 0 -4 = -4$

$x = 1 , y = 2(1) -4 = 2 -4 = -2$

$x = 2, y =2(2) - 4 = 4-4 = 0$

$x$

$0$

$1$

$2$

$3$

$4$

$y$

$- 4$

$-2$

$0$

$2$

$4$

$(ii)$

$E$

$F$

$G$

$H$

$I$

triangle formed by the lines with y -axis

$\Delta AEC$ coordinates of vertices are

$A(0,4), E(0 , -4)$ and

$C ( 2, 0 )$

Area of

$\Delta AEC = \dfrac {1}{2} \quad Base \times Altitude$

$= \dfrac {1}{2} AE \times CO$

$= \dfrac {1}{2} \times [ 4 -(-4)] \times ( 2- 0 )$

$= \dfrac {1}{2} \times 8 \times 2 = 8$

$\therefore$ Area of

$\Delta AEC =$ Square units.

1656287_1795687_ans_085a1105bdca490bb16696706de8d375.PNG

Find the solution of the pair of equations $\frac {x}{10}+ \frac {y}{5} -1 = 0$ and $\frac {x}{8} +\frac {y}{6} = 15$ hence find $\lambda$ if y = $\lambda x +5$

given equations are

$\dfrac {x}{10} + \dfrac {y}{5} - 1 = 0 ...(i) \times 20$

And $\dfrac {x}{8} + \dfrac {y}{6} = 15 ...(ii) \times 24$

$2x + 4y = 20 ...(iii)$

$3x +4y = 360 ... (iv )$

$2x +4y = 20 ...(iii)$

$3x +4y = 360 ...(iv)$

$- \quad - \quad -$ [ subtracting $(iv)$ from $(iii)$ ]

_______________________

$-x \quad = -340$

$\Rightarrow 4x = +340$

now $2x +4y =20$ [ from (iii) ]

$\Rightarrow x +2y = 10$

$\Rightarrow 340 +2y = 10 [ Rs x = 340 ]$

$\Rightarrow 2y = 10 -340$

$\Rightarrow 2y = -330$

$\Rightarrow y = \dfrac {-330}{2} = -165$

and $x =340$

Now , $y = \lambda x + 5$ [ given ]

$\Rightarrow - 165 = \lambda ( 340) + 5 [Rs y = -165$ and $x =340$ ]

$\Rightarrow -\lambda (340) = 170$

$\Rightarrow \lambda = \dfrac {170}{-340} \Rightarrow \lambda = - \dfrac {1}{2}$

hence the solution of the given pair of equation is $x = 340 , y = -165$ and $\lambda = - \dfrac {1}{2}$ .

A shopkeeper given books on rent for reading .She takes a fixed charge for the first two days and an additional charge for each day thereafter. Latika paid $Rs 22$ for a book kept for six days, while anand paid $rs 16$ for the book kept for days . find the fixed charges and the charge for each extra day.

Let the fixed charges for first two days

$= Rs x$
let the additional charges per day after

$2$ days =

$rs y$
Latika paid

$rs 22$ for six days [given ]

$2$ days fixed charges

$+( 6-2 )$ days charges = 22

$\Rightarrow x + 4y = 22$
Anand paid

$rs 16$ for books kept for four days.

$2$ day's fixed charges

$+( 4- 2)$ day's additional charges

$= 16$

$\Rightarrow x +2y =16.....(ii)$

$x + 2y = 16 [ from (ii) ]$

$x +4y = 22 [ from (i) ]$

$- \quad -\quad -$
___________________

$-2y = -6$ [ Subtracting eqn.

$(i)$ form

$(ii)$ ]

$\Rightarrow y = Rs 3$ per day
Now

$X + 2y = 16$ [ from (i) ]

$\Rightarrow x + 2(3)= 16$

$\Rightarrow x = 16 -6 \Rightarrow x = 10$
So the fixed charges for first

$2$ days =

$rs 10$
The additional charges per day after

$2$ days =

$Rs 3$ per day

Solve the following pairs of linear equations :

$43x +67 y = -24 , 67x +43y =24$

Given pair of equations are

$43x +67y = -24 ...(i)$

$67x + 43y = 24 ....(ii)$

$\Rightarrow 110x +110y =0$ [ Adding

$(i)$ and

$(ii)$ ]

$\Rightarrow x +y = 0 ...(iii)$
Subtracting

$(ii)$ from

$(i)$ we have

$-24 x +24 y = -48$

$\Rightarrow -x +y = -2 ....(iv)$

$x +y = 0$ [ From (iii) ]

$\Rightarrow 2y = -2$ [ adding

$(iii)$ and

$(iv)$ ]

$y = -1$
From

$(iii) , x+y = 0$

$\Rightarrow x +(-1) = 0 [ \because y = -1 ]$

$\Rightarrow x = 1$
and

$y = -1$
hence the solution of the given equations is

$x = 1, y = -1$

Graphically , solve the following pair of equations $2x +y = 6$ and $2x-y +2 = 0$ find the ratio of the areas of the two triangles formed by the lines representing these equations with equations with the $x-axis$ and the lines with the $y -axis$

Given equation is

$2x + y = 6$

$y = 6 -2x ....(i)$

$x = 0 , y = 6 - 2 (0) = 6$

$x = 1 , y = 6 -2 (1) = 6 -2 = 4$

$x = 2 , y = 6 -2(2) = 6-4 = 2$

$x$

$0$

$1$

$2$

$y$

$6$

$4$

$2$

$I$

$A$

$B$

$C$

given equation is

$2x - y + 2 = 0$

$\Rightarrow y = 2x +2 ....(ii)$

$x = 0 , y = 2(0) = 0 +2 = 2$

$x = , y = 2(1) +2 = 2 +2 = 4$

$x = 2, y = 2(2) + 2 = 4 +2 = 6$

$X$

$0$

$1$

$2$

$y$

$2$

$4$

$6$

$II$

$D$

$E$

$F$

The area of

$\Delta BGH$ formed by lines and x axis

$= \dfrac {1}{2} GH \times BJ$

$= \dfrac {1}{2} [ 3 -(-1) \times ( 4- 0) = \dfrac {1}{2} \times 4 \times 4 = 8$ sq units.

the area of

$\Delta BAD$ formed by lines and Y axis

$= \dfrac {1}{2} AD \times KB$

$= \dfrac {1}{2} [ ( 6 -2) \times ( 1-0 ) = \dfrac {1}{2} \times 4 \times 1 = 2$ sq units

$\therefore$ ratio of areas of two

$\Delta s \dfrac {Area \Delta BGH }{ Area \Delta BAD } = \dfrac {8}{2} = \dfrac {4}{1} = 4 : 1$

1656223_1795724_ans_fdc201c680c947be983fd8384b391878.png

By the graphical method, find the whether the following pair of equation are consistent or not . if consistent solve them.

$x-2y = 6$ and $3x -6y = 0$

Given equations are,

$x -2y = 6 ...(i)$

$3x - 6y = 0....(ii)$

$\dfrac {a_1}{a_2} = \dfrac {1}{3} , \dfrac {b_1}{b_2} = \dfrac {-2}{-6} = \dfrac {1}{3} , \dfrac {c_1}{c_2} = \dfrac {6}{0}$

$\therefore \dfrac {a_1}{a_2} = \dfrac {b_1}{b_2} \neq \dfrac {c_1}{c_2}$

$\therefore$ System of equations is inconsistent , hence the lines represented by the given equations are parallel . so the given equations have no solutions

Draw the graphs of the equations $x = 3, x = 5$ and $2x -y-4 = 0$ also find area of the quadrilateral formed by the lines and the x-axis.

x	0	1	2	3	4
y	-4	-2	0	2	4
III	G	H	J	K	L

The given equations are

$x = 3 ....(i)$

$x = 5 .....(ii)$

$2x - y - 4 = 0$

$\Rightarrow y = 2x - 4$

$x$	$3$	$3$	$3$	$x$	$5$	$5$	$5$
$y$	$1$	$2$	$3$	$y$	$3$	$4$	$5$
$I$	$A$	$B$	$C$	$II$	$D$	$E$	$F$

The coordinates of the vertices of the required

$rs PMNB$ are

$P (3,0), M (5, 0 ) ,N( 5 , 6)$ and

$B ( 3 , 2)$

1656075_1795729_ans_a29314b678bf49dcb33bd8bcfc7e4952.png

Solve the following pair of linear equations by any suitable method:
$x+y=5$
$2 x-3 y=5$

Given,

$x+y=5$

$x=5-y$

Substituting the value of

$x$ in

$2 x-3 y=5$

$2(5-y)-3 y=5$

$10-2 y-3 y=5$

$5=5 y$

$y=1$

Substituting the value of

$y$ in

$x=5-y$

$x=5- y$

$x=5-1$

$x=4$

Solve the following simultaneous linear equation
$x+y=5.5$ and $x-y=0.9$

Given equations are,

$x+y=5.5$ ...(i)

$x-y=0.9$ ...(ii)

Adding eq(i) and eq(ii), we get

$2x=5.5 +0.9$

$\Rightarrow x=\dfrac{6.4}2$

$\Rightarrow x=3.2$

Substituting value of

$x$ in eq(i)

$3.2 +y=5.5$

$\Rightarrow y=5.5 -3.2$

$\Rightarrow y=2.3$

Hence

$x=3.2$ and

$y=2.3$

A motor boat can travel $30$ km upstream and $28$ km downstream in $7$ hours , it can travel $21$ km upstream and return in $5 hours$ find the speed of the boat in still water and the speed of the stream.

Let speed of boat in still water

$= x \ km/hr$

and the speed of the stream

$= y \ km/hr$

Speed of motor boat upstream

$= (x - y) km/hr$

Speed of motor boat downstream

$= (x + y) km/hr$

Case I : Time taken by motor boat in

$30 km$ upstream

$= \dfrac{30}{x -y} hr$

Time taken by motor boat in

$28 km$ downstream

$= \dfrac{28}{x + y } hr$

$\therefore \dfrac{30}{(x - y)} + \dfrac{28}{(x + y)} = 7$

$\Rightarrow 2 \left[\dfrac{15}{(x - y)} + \dfrac{14}{(x + y)} \right] = 7$

$\Rightarrow \dfrac{15}{x - y} + \dfrac{14}{x + y} = \dfrac{7}{2}$ ___(i)

Case II : Time taken by motor boat in

$21 km$ upstream

$= \dfrac{21}{x - y} hr$

Time taken by motor boat to return

$21 km$ downstream

$= \dfrac{21}{x + y} hr$

$\therefore \dfrac{21}{x - y} + \dfrac{21}{x + y} = 5$

$\Rightarrow 21 \left[\dfrac{1}{x - y} + \dfrac{1}{x + y} \right] = 5$

$\Rightarrow \dfrac{1}{x - y} + \dfrac{1}{x + y} = \dfrac{5}{21}$ ____(ii)

$\dfrac{15}{x - y} + \dfrac{14}{x + y } = \dfrac{7}{2}$ [From (i)]

As equations (both) are symmetric to

$(x - y)$ and

$(x + y)$ so we can eliminate either

$(x - y)$ or

$(x + y)$

Multiplying (ii) by

$14$ , we get

$\dfrac{14}{(x - y)} + \dfrac{14}{(x + y)} = \dfrac{70}{21}$ ___(iii)

$\dfrac{15}{(x - y)} + \dfrac{14}{x + y} = \dfrac{7}{2}$ [From (i)]

$-$

$\overline{\dfrac{14}{(x - y)} - \dfrac{15}{(x - y)} = \dfrac{10}{3} - \dfrac{7}{2}}$ [Subtracting (i) from (iii)]

$\Rightarrow \dfrac{14 - 15}{(x - y)} = \dfrac{20 - 7 \times 3}{3 \times 2}$

$\Rightarrow \dfrac{-1}{(x - y)} = \dfrac{-1}{6}$ ___(iv)

$\Rightarrow (x - y) = 6$

Now, substituting

$x - y = 6$ in (ii), we have

$\dfrac{1}{(x - y)} + \dfrac{1}{(x + y)} = \dfrac{5}{21}$

$\Rightarrow \dfrac{1}{6} + \dfrac{1}{(x + y)}= \dfrac{5}{21}$

$\Rightarrow \dfrac{1}{(x + y)} = \dfrac{5}{21} - \dfrac{1}{6}$

$\Rightarrow \dfrac{1}{(x + y)} = \dfrac{2 \times 5 - 7 \times 1}{3 \times 7 \times 2}$

$\Rightarrow \dfrac{1}{(x + y)} = \dfrac{3}{42}$

$\Rightarrow \dfrac{1}{(x + y)} = \dfrac{1}{14}$

$\Rightarrow x + y = 14$ __(v)

$x - y = 6$ [From (iv)]

$\overline{2x \,\,\,\,\,= 20}$ [Subtracting (iv) from (v)]

$\Rightarrow x = 10 km/hr$

Now,

$x + y = 14$ [from (v)]

$\Rightarrow 10 + y = 14$

$\Rightarrow y = 4 km/hr$

Hence, the speed of motorboat and stream are

$10 km/hr$ and

$4 km/hr$ respectively.

Solve the following simultaneous linear equation
$x+y=7xy$ and $2x-3y+xy=0$

Given equations are,

$x+y=7xy$ ...(i)

$2x-3y+xy=0$ ...(ii)

Dividing eq (i) by

$xy$ , we get

$\dfrac{x}{xy}+\dfrac{y}{xy}=\dfrac{7xy}{xy}$

$\Rightarrow \dfrac{1}{y}+\dfrac{1}{x}=7$

$\Rightarrow \dfrac{1}{x}+\dfrac{1}{y}=7$ ...(iii)

Divide eq (ii) by

$xy$ , we get

$\dfrac{2x}{xy}-\dfrac{3y}{xy}+\dfrac{xy}{xy}=0$

$\Rightarrow \dfrac{2}{y}-\dfrac{3}{x}+1=0$

$\Rightarrow -\dfrac{3}{x}+\dfrac{2}{y}=-1$ ...(iv)

Multiply eq (iii) by 3, we get

$\dfrac{3}{x}+\dfrac{3}{y}=3\times 7$

$\Rightarrow \dfrac{3}{x}+\dfrac{3}{y}=21$ ...(v)

Adding eq (v) and eq (iv), we get

$\dfrac{5}{y}=20$

$\Rightarrow y=\dfrac{5}{20}$

$\Rightarrow y=\dfrac{1}{4}$

Substituting the value of

$y$ in eq (iv), we get

$-\dfrac{3}{x}+2\times 4=-1$

$\Rightarrow -\dfrac{3}{x}+8=-1$

$\Rightarrow -\dfrac{3}{x}=-1-8$

$\Rightarrow -\dfrac{3}{x}=-9$

$\Rightarrow x=\dfrac{3}{9}$

$\Rightarrow x=\dfrac{1}{3}$

Hence,

$x=\dfrac 13$ and

$y=\dfrac 14$ .

Determine , algebraically , the vertices of the triangle formed by the lines $3x -y =3 , 2x -3y = 2$ and $x +2y = 8$ .

Given linear equations are

$3x - y = 3$ ___(i)

$2x - 3y = 2$ ___(ii)

$x + 2y = 8$ ___(iii)

Let the intersecting points of lines (i) and (ii) is A, and of lines (ii) and (iii) is B and that of lines (iii) and (i) is C.

The intersecting point of (ii) and (i) can be find out by solving (i) and (ii) for (x, y).

$3x - y = 3$ [From (i)]

$2x - 3y = 2$ [From (ii)]

$9x - 3y = 9$ __(iv) [multiplying eqn. (i) by 3]

$2x - 3y = 2$ [from (ii)]

$-$

$+$

$-$

$\overline{7x\,\,\,\,\,\,\,\,\,\,= 7}$ [By subtracting (ii) from (iv)]

$\Rightarrow x = \dfrac{7}{7} \Rightarrow x = 1$

Now,

$3x - y = 3$ [From (i)]

$\Rightarrow 3(1) - y = 3 \, [x = 1]$

$\Rightarrow -y = 3 - 3$

$\Rightarrow -y = 0$

$\Rightarrow y = 0$

So, intersecting point of eqns.(i) and (ii) is

$A (1, 0)$ .

Similarly, intersecting point B of eqns. (ii) and (iii) can be find out as follows:

$2x - 3y = 2$ [from (ii)]

$x + 2y = 8$ [from (iii)]

$2x - 3y = 2$ [From (ii)]

$2x + 4y = 16$ __(v) [By multiplying (iii) by 2]

$-$

$\overline{\,\,\,-7y = -14}$ [Subtracting (v) from (ii)]

$\Rightarrow y = \dfrac{14}{7} \Rightarrow y = 2$

Now,

$x + 2y = 8$ [From (iii)]

$\Rightarrow x + 2 (2) =8$

$\Rightarrow x = 8 - 4$

$\Rightarrow x = 4$

So, the coordinates of B are

$(4, 2)$

Similarly, for intersecting point of C of eqns. (i) and (iii), we have

$3x - y = 3$ [From (i)]

$x + 2y = 8$ [From (iii)]

Multiplying (i) by 2, we get

$6x - 2y = 6$ ___(vi)

$x + 2y = 8$ [From (iii)]

$\overline{7x \,\,\,\,\,\,\,\, = 14}$ [Adding (vi) and (iii)]

$\Rightarrow x = \dfrac{14}{7} \Rightarrow x = 2$

Now,

$3x - y = 3$ [from (i)]

$\Rightarrow 3(2) - y = 3$

$\Rightarrow -y = 3 - 6$

$-y = -3 \Rightarrow y = 3$

So, point C is

$(2, 3)$

Hence, the vertices of

$\Delta ABC$ formed by given three linear equations are

$A(1,0), B (4, 2)$ and

$C(2, 3)$

The cost of $4$ pens and $4$ pencil boxes is $Rs 100$ . Three times the cost of a pen is $Rs 15$ more than the cost of pencil box. from the pair of linear equations for the above situation. find the cost of a pen and a pencil box.

Let the cost of a pen =

$rs x$
Let the cost of a pencil box =

$Rs y$

$\therefore$ the cost

$4$ pencils boxes

$Rs 100$ [given ]

$4x +4y= 100....(i)$

$x + y = 25 ....(ii)$ [ by dividing (i) by

$4$ ]
according to the second condition we have,

$3x = y + 15$

$3x -y = 15 .....(iii)$

$x +y = 25 ....(ii)$
________________

$4x =40$ [adding (ii) and (iii) ]

$\Rightarrow x = \frac { 40}{ 4} = 10$
Now

$x + y = 25$ [ from (i) ]

$10 +y = 25 [ Rs x = 10 ]$

$\Rightarrow y = 25 -10 = rs 15$
So

$x = rs 10$ and

$y = rs 15$
Hence , the cost of pen and a pencil box are

$Rs 10$ and

$Rs 15$ respectively.

A railway half ticket costs half the full fare, but the reservation charges are the same on a half ticket as on a full ticket . one reserved first class ticket from the station $A$ to $B$ cost $rs 2530$ . Also On reserved first class ticket and one reserved first class half ticket from $A$ to $B$ cost $rs 3810$ . Find the full first class fare from station $A$ to $B$ , and also the reservation charges for a ticket.

$\textbf{Step-1: Assume the required unknown to be equal to some variables}$

$\textbf{& then apply all information's given.}$

$\text{Let the reservation charges be Rs.}$

$x$

$\text{and cost of full ticket of first class be Rs. y}$

$\text{It is given that one reserved first class ticket from the station}$

$\text{A to B costs Rs. 2530. Also, one reserved first class}$

$\text{ticket and one reserved first class half ticket from A to B costs Rs. 3810, therefore,}$

$x+y=2530 ..........(1)$

$\text{and}$

$\text{since railway half ticket costs half the full fare, but the reservation charges are the same}$

$\text{on a half ticket as on a full ticket. So, we get,}$

$(x+y)+\left( x+\frac { y }{ 2 } \right) =3810\quad \\ \Rightarrow 4x+3y=7620.............(2)$

$\textbf{Step-2: Simplify unknown.}$

$\text{Multiplying equation 1 by}$

$3,$

$\text{we get,}$

$3x+3y=7590 ..........(3)$

$\text{Now subtracting equation 3 from equation 2, we get}$

$x=30$ .

$\text{Now substitute the value of}$

$x$

$\text{in equation 1 that is:}$

$30+y=2530$

$\Rightarrow$

$y=2500$

$\textbf{Hence, the full first class fare from station A to B is Rs. 2500}$

$\textbf{& the reservation charges for a ticket is Rs. 30}$

Akita travels $14$ km to her home partly by rickshaw and partly by bus. she takes half an hour if she travels $2$ km by rickshaw and the remaining distance by bus . On the other hand , if she travels $4$ km by rickshaw and the remaining distance by bus, she takes $9$ minutes longer. find the speed of the rickshaw and of the bus.

Let the speed of rikshaw $= x km / hr$

and let the speed of bus = $y lm/hr$

case I : time taken by rickshaw to travel $2km = \dfrac {Distance}{Speed} = \dfrac {2}{x}$

time taken by bus to travel $( 14-2 ) km$ ( remaining ) = $\dfrac { 12}{ y}$ hr

Total times taken by rickshaw $(2 km)$ and bus $( 12 km ) = \dfrac {1}{2} hr$

$\therefore \dfrac {2}{x} + \ frac { 12}{y} = \dfrac {1}{2} ...(i)$

Case II : Time taken by rickshaw to travel $4 km = \dfrac { 4}{x} hr$

Time taken by bus of travel remaining $(14 - 4) km = \dfrac {10}{ y} hr$

$\therefore$ Total time in case $II$ = $\dfrac { 1}{2} hr + 9 min$

$\therefore \dfrac {4}{x} + \dfrac {10}{y} = \dfrac {1}{2} hr + \dfrac {9}{60} hr$

$\Rightarrow \dfrac {4}{x} + \dfrac { 10 }{y } = \dfrac { 30 + 9}{60}$

$\Rightarrow \dfrac {4}{x} + \dfrac {10}{y} = \dfrac { 39}{60}$

$\Rightarrow \dfrac {4}{x} + \dfrac {10}{y} = \dfrac {13}{20} ....(iii)$

Multiplying equation (i) by $2$ we get

$\dfrac {4}{x} + \dfrac {24}{y} = \dfrac {2}{ 2}$ .....(iii)

Now subtracting (iii) from (ii) we get

$\dfrac {4}{x} + \dfrac {10}{y} = \dfrac {13}{20}$

$\dfrac {4}{x} +\dfrac {24}{y} = \dfrac {2}{2}$

$- \quad - \quad -$

_______________________

$\dfrac {10}{y} - \dfrac {24}{y} = \dfrac {13}{20} - \dfrac {2}{2}$

$\Rightarrow \dfrac {10-24}{y} = \dfrac {13-20}{20} \Rightarrow - \dfrac {14}{y} = \dfrac {-7}{20}$

$\Rightarrow 7y = 14 \times 20$

$\Rightarrow y = \dfrac {14 \times 20}{7}$

$\Rightarrow y = 40 km/hr$

$Now , \dfrac {2}{x} + \dfrac {12}{y} = \dfrac {1}{2} [from (i) ]$

$\Rightarrow \dfrac {2}{x} + \dfrac {12}{(40) } = \dfrac {1}{2}$

$\Rightarrow \dfrac {2}{x} = \dfrac {1}{2} - \dfrac {3}{10}$

$\Rightarrow \dfrac {2}{x} = \dfrac { 5-3}{10}$

$\Rightarrow \dfrac {2}{x} = \dfrac {2}{10}$

$\Rightarrow x = 10 km/hr$

hence the speeds of rickshaw and bus are $10 km/hr$ and $40km/ hr$ respectively.

Determine , graphically , the vertices of the triangle formed by the lines $y = x, 3y = x, x +y = 8$ .

Given equation are

$y = x ...(i)$

$x = 3y ....(ii)$

$x +y = 8 ....(iii)$

$\Rightarrow y = 8 -x [ from (iii) ]$

$x = 0 , y = 8 -0 = 8$

$x = 1 , y = 8 -1 = 7$

$x 2 ,y = 8 -2 = 6$

$y = 8 -x$

$III$

$J$

$K$

$L$

$M$

$x$

$0$

$1$

$2$

$8$

$y$

$8$

$7$

$6$

$0$

$y = x$

$I$

$A$

$B$

$C$

$D$

$X$

$0$

$1$

$2$

$3$

$y$

$0$

$1$

$2$

$3$

$x = 3y$

$II$

$E$

$F$

$G$

$H$

$x$

$0$

$1$

$2$

$3$

$y$

$0$

$3$

$6$

$9$

Hence the vertices of

$\Delta GNA$ formed by

$3$ lines are

$G ( 2, 6 ) , N( 4 , 4 )$ and

$A ( 0 , 0)$ .

1656086_1795726_ans_9a8ebc0610c749b3ba9a9ca36c7a219a.png

The cost of $5$ kg of sugar and $7$ kg of rice is Rs. $153$ , and the cost of $7$ kg of sugar and $5$ kg of rice is Rs. $147$ . Find the cost of $6$ kg of sugar and $10$ kg of rice.

Let’s assume the cost of

$1$ kg of sugar

$=$ Rs

$x$

And, let the cost of

$1$ kg of rice

$=$ Rs

$y$

Then according to the given conditions, we have

$5x + 7y = 153$ … (i) and

$7x + 5y = 147$ … (ii)

Multiplying (i) by

$7$ and (ii) by

$5$ , we have

$35x + 49y = 1071$ … (iii)

$35x + 25y = 735$ … (iv)

Subtracting (iv) from (iii), we get

$24y = 336$

$\Rightarrow y = \dfrac{336}{24}$

$\Rightarrow y = 14$

On substituting the value of

$y$ in (i), we get

$5x + 7(14) = 153$

$\Rightarrow5x + 98 = 153$

$\Rightarrow 5x = 153 - 98$

$\Rightarrow 5x = 55$

$\Rightarrow x = \dfrac{55}5 = 11$

Hence,

The cost of

$1$ kg of sugar is Rs

$11$ and

The cost of

$1$ kg of rice is Rs

$14$ .

Now,

Cost of 6 kg of sugar

$=$ Rs

$11 \times 6 =$ Rs

$66$

Cost of 10 kg of rice

$=$ Rs

$14 \times 10 =$ Rs

$140$

Thus, the cost of

$6$ kg of sugar and

$10$ kg of rice

$=$ Rs

$66 +$ Rs

$140 =$ Rs

$206.$

The class IX students of a certain public school wanted to give a farewell party to the outgoing students of class X. They decided to purchase two kinds of sweets, one costing Rs. $70$ per kg and the other costing Rs. $84$ per kg. They estimated that $36$ kg of sweets were needed. If the total money spent on sweets was Rs. $2800$ , find how much sweets of each kind they purchased.

Let

The quantity of sweet costing Rs

$70$ be

$x$ and,

The quantity of sweet costing Rs

$84$ be

$y$

Given, total quantity of sweets purchased is

$36$ kg

i.e.,

$x + y = 36$ … (i)

Also given, the total money spent is Rs

$2800$

i.e.,

$70x + 84y = 2800$ … (ii)

Multiplying (i) by

$70$ , we get

$70x + 70y = 2520$ ... (iii)

Subtacting (ii) from (iii), we get

$-14y = -280$

$\Rightarrow y = \dfrac{-280}{-14}$

$\Rightarrow y = 20$

On substituting the value of

$y$ in equation (i), we get

$x + 20 = 36$

$\Rightarrow x = 36 – 20$

$\Rightarrow x = 16$

Therefore, the quantities of sweets purchased are

$16$ kg which costs Rs

$70$ per kg and

$20$ kg which costs Rs

$84$ per kg.

The sum of two numbers is $43$ . If the larger is doubled and the smaller is tripled, the difference is $36$ . Find the two numbers.

Let’s assume the two numbers to be

$x$ and

$y$ such that

$x > y$

Then according to the given conditions, we have

$x + y = 43$ … (i) and

$2x - 3y = 36$ … (ii)

Now, multiplying (i) by

$3$ and adding with (ii) we get

$3x + 3y = 129$

$2x - 3y = 36$

$\overline {\ \ 5x\ \ \ = 165\ \ \ }$

$\Rightarrow x = \dfrac{165}5 = 33$

On substituting the value of

$x$ in (i), we get

$33 + y = 43$

$\Rightarrow y = 43 – 33$

$\Rightarrow y = 10$

Therefore, the two numbers are

$33$ and

$10$ .

$2$ men and $5$ women can do a piece of work in $4$ days, while one man and one woman can finish it in $12$ days. How long would it take for $1$ man to do the work?

Let’s assume that a man takes

$x$ days to do work and a woman takes

$y$ days.

So, the amount of work done by 1 man in 1 day

$= \dfrac 1x$

And the amount of work done by 1 woman in 1 day

$= \dfrac 1y$

Now,

The amount of work done by

$2$ men in 1 day will be

$= \dfrac 2x$

And the amount of work done by

$5$ women in 1 day

$= \dfrac5y$

Then according to the question,

$\dfrac 2x + \dfrac5y = \dfrac{1}{4}$ … (i)

$\dfrac 1x + \dfrac 1y = \dfrac1{12}$ … (ii)

Multiplying equation (ii) by

$5$ , we get

$\dfrac 5x + \dfrac 5y = \dfrac 5{12}$ ... (iii)

Subtracting equation (i) from (iii), we get

$\dfrac 3x = \dfrac 5{12} - \dfrac{1}{4}$

$\Rightarrow \dfrac 3x = \dfrac{5 - 3}{12}$

$\Rightarrow \dfrac 3x = \dfrac {2}{12} = \dfrac 16$

$\Rightarrow x = 18$

Therefore,

$1$ man can do the work in

$18$ days.

There are $38$ coins in a collection of $20$ paise coins and $25$ paise coins. If the total value of the collection is Rs. $8.50$ , how many of each are there?

${\textbf{Step -1: Illustrate the data and apply the condition. }}$

${\text{Let the total number of 20 paise coins be a}}$

${\text{and the total number of 25 paise coins be b}}{\text{.}}$

${\text{Now, accoring to the question}}$

$a + b = 38{\text{ }} - - - \left( i \right)$

${\text{and, }}20a + 25b = Rs.850$

$\Rightarrow 20a + 25b = 850paise{\text{ }} - - - \left( {ii} \right)$

$\boldsymbol{{\textbf{Step -2: Solve equation }}\left( i \right){\textbf{ and }}\left( {ii} \right.)}$

${\text{Apply, }}\left\{ {\left( {ii} \right) - 20 \times \left( i \right)} \right\}$

$\therefore {\text{ }}\left( {20a + 25b} \right) - \left( {20a + 20b} \right) = 850 - \left( {20 \times 38} \right)$

$\Rightarrow 5b = 90$

$\Rightarrow b = 18$

${\text{Substitute the value in equation }}\left( i \right)$

$\therefore a + 18 = 38$

$\Rightarrow a = 20$

${\text{Thus, the number of 20 paise coins are 20 and the number of 25 paise coins are 18}}$

${\textbf{Hence, the number of 20 paise coins are 20 and the number of 25 paise coins are 18}}{\textbf{.}}$

Solve the following simultaneous linear equation
$3x+2y=2xy$

$\dfrac{1}{x}+\dfrac{2}{y}=1\dfrac{1}{6}$

Given equations are,

$3x+2y=2xy$

$\Rightarrow \dfrac{3x}{xy}+\dfrac{2y}{xy}=\dfrac{2xy}{xy}\qquad\text{[ dividing by } xy\ ]$

$\Rightarrow \dfrac{3}{y}+\dfrac{2}{x}=2$ ...(i)

And,

$\dfrac{1}{x}+\dfrac{2}{y}=1\dfrac{1}{6}$

$\Rightarrow \dfrac{1}{x}+\dfrac{2}{y}=\dfrac{7}{6}$ ...(ii)

Multiplying eq(ii) by 2, we get

$\dfrac{2}{x}+\dfrac{4}{y}=\dfrac{7}{3}$ ...(iii)

Subtracting eq (i) from eq(iii), we get,

$\ \left(\dfrac{2}{x}+\dfrac{4}{y}\right)-\left( \dfrac{3}{y}+\dfrac{2}{x} \right)=\dfrac{7}{3}-2$

$\Rightarrow \dfrac{1}{y}=\dfrac{7-6}{3}$

$\Rightarrow \dfrac{1}{y}=\dfrac{1}{3}$

$\Rightarrow y=3$

Substituting the value of

$y$ in eq(i), we get

$\dfrac 33+\dfrac 2x=2$

$\Rightarrow 1+\dfrac 2x=2$

$\Rightarrow 2x=1$

$\therefore x=2$

Hence

$x=2$ and

$y=3$

Solve the following simultaneous linear equation
$ax+by=a-b$ and $bx-ay=a+b$

Given equations are,

$ax+by=a-b$ ...(i)

$bx-ay=a+b$ ...(ii)

Multiplying eq (i) by

$a$ and eq (ii) by

$b$ , we get

$a(ax+by)=a(a-b)$

$\Rightarrow a^2x+aby=a^2-ab$ ...(iii)

$b(bx-ay)=b(a+b)$

$\Rightarrow b^2x-aby=ab+b^2$ ...(iv)

Adding eq (iii) and eq(iv)

$\ \ \ a^2x+aby=a^2-ab$

$\ \ \ b^2x-aby=ab+b^2$

$\overline{(a^2+b^2)x=(a^2+b^2)}$

$\Rightarrow x=\dfrac{(a^2+b^2)}{(a^2+b^2)}$

$\Rightarrow x=1$

Substituting the value of

$x$ in eq (i), we get

$a \times 1+by=a-b$

$\Rightarrow a+by=a-b$

$\Rightarrow by=-b$

$\Rightarrow y=-\dfrac bb$

$\Rightarrow y=-1$

Hence

$x=1$ and

$y=-1$ .

A shopkeeper sells a table at $8\%$ profit and a chair at $10\%$ discount, thereby getting Rs. $1008$ . If he had sold the table at $10\%$ profit and chair at $8\%$ discount, he would have got Rs. $20$ more. Find the cost price of the table and the list price of the chair.

Let C.P. of table

$=$ Rs

$x$ and

C.P. of chair

$=$ Rs

$y$

The shopkeeper sells the table at

$8\%$ profit and the chair at

$10\%$ discount

$\Rightarrow x \times \dfrac{ 108}{100}+y \times \dfrac{ 90}{100}=1008$

$\Rightarrow 108x + 90y = 100800$

$\Rightarrow 6x + 5y = 5600$

$… (i)$

The shopkeeper sells the table at

$10\%$ profit and the chair at

$8\%$ discount and got

$Rs.20$ more.

$\Rightarrow x\times \dfrac{110}{100}+y \times \dfrac{ 92}{100}=1008+20$

$\Rightarrow 110x + 92y = 102800$

$\Rightarrow 55x + 46y = 51400$

$… (ii)$

Now, multiplying

$(i)$ by

$55$ and

$(ii)$ by

$6,$ we have

$330x + 275y = 308000$

$... (iii)$

$330x + 276y = 308400$

$... (iv)$

Subtracting equation

$(iv)$ from

$(iii),$ we get

$\Rightarrow -y = -400$

$\Rightarrow y = 400$

On substituting the value of

$y$ in equation

$(i),$ we get

$\Rightarrow 6x + 5(400) = 5600$

$\Rightarrow 6x = 5600 - 2000$

$\Rightarrow x = \dfrac{3600}6$

$\Rightarrow x = 600$

Hence, Cost Price of the table

$=$ Rs

$600$

List price of chair

$=400\times \dfrac{90}{100}=$ Rs

$360$

Solve the following pair of equations graphically. Plot at least $3$ points for each straight line $2x -7y = 6, 5x -8y = -4.$

$2x-7y = 6$ ….(i)

$2x = 7y+6$

$x = (7y+6)/2$
when

$y = 0$

$x = (7×0+6)/2 = 6/2 = 3$
when

$y = -1$

$x = (7×-1+6)/2 = -1/2 = -0.5$
when

$y = -2$

$x = (7×-2+6)/2 = -8/2 = -4$

$x=$	$3$	$,-0.5$	$,-4$
$y=$	$0$	$,-1$	$,-2$

Mark the above points on graph. Join them.

$5x-8y = -4$ …(ii)

$5x = 8y-4$

$x = (8y-4)/5$
when

$y = 0$

$x = (8×0-4)/5 = -4/5 = 0.8$
when

$y = 3$

$x = (8×3-4)/5 = (24-4)/5 = 20/5 = 4$
when

$y = -2$

$x = (8×-2-4)/5 = (-16-4)/5 = -20/5 = -4$

$x=$	$0.8$	$,4$	$,-4$
$y=$	$0$	$,3$	$,-2$

Mark the above points on graph. Join them.
It is clear from the graph that the two lines intersect at

$(-4,-2).$
So the solution of the given equations are

$x = -4$ and

$y = -2.$

1725074_1812145_ans_146bf3f4086545968196d1d0b1bb7647.PNG

Pair Of Linear Equations In Two Variables - Class 10 Maths - Extra Questions

Solve the following equations graphically: 3x−2y=43x-2y = 4, 5x−2y=05x-2y = 0

Solve the following pairs of linear equations by substition method;3x−y=33x - y = 39x−3y=99x - 3y = 9

Solve the following pair of linear equations by substitution method:x+y=2mx + y = 2mmx−ny=m2+n2mx - ny = m^2+n^2

Solve the following pair of linear equations by substitution method:x2+y=0.8\dfrac{x}{2} + y = 0.8x+y2=710x + \dfrac{y}{2} = \dfrac{7}{10}

Solve the following pair of linear equations by substitution method:xa+yb=2,a≠0,b≠0\dfrac{x}{a} +\dfrac{y}{b} = 2 , a\neq 0,b \neq 0ax−by=a2−b2ax - by = a^2 - b^2

Solve graphically the following system of linear equations if it has unique solution :3x+y=23x + y = 26x+2y=16x + 2y = 1

Find graphically the co-ordinates of the vertices of the triangle formed by the lines y=0,y=xy = 0, y = x and 2x+3y=102x + 3y= 10. Hence find the area of the triangle formed by these lines.

Solve the following pair of linear equations by substitution method:0.5x - 0.8y = 3.40.6x + 0.3y = 0.3

Solve the following pair of linear equations by substitution method:S−t=3S - t = 3s3+t2=6\dfrac{s}{3} + \dfrac{t}{2} = 6

Using the same axes of co-ordinates and the same unit, solve graphically.x+y=0,3x2y=10x+y = 0, 3x 2y = 10

Solve the following pair of linear equations by substitution method:7x−15y=27x - 15y = 2x+2y=3x + 2y = 3

Solve the following systems equations by eliminations method;3x−5y−4=0...(i)3x - 5y - 4 = 0...(i)9x=2y+7....(ii)9x = 2y + 7....(ii)

Solve the following systems equations by eliminations method;8x+5y=98x + 5y = 93x+2y=4 3x + 2y = 4

Solve the following systems equations by eliminations method;0.4x+1.5y=6.50.4x + 1.5y = 6.50.3x−0.2y=0.9 0.3x - 0.2y = 0.9

Solve the following systems equations by eliminations method;√2x−√3y=0\sqrt{2}x - \sqrt{3}y = 0√5x+√.2y=0 \sqrt{5}x + \sqrt{.2}y = 0

Solve the following systems equations by eliminations method;2x+3y=82x + 3y = 84x+6y=74x + 6y =7

Solve the following pair of linear equations by substitution method:bxa+ayb=a2+b2\dfrac{bx}{a} + \dfrac{ay}{b} = a^2 + b^2x+y=2abx + y = 2ab

Solve graphically the following system of linear equations if is has unique solution:2x−6y+10=02x - 6y + 10 = 03x−9y+15=03x - 9y + 15 =0

Solve the following systems equations by eliminations method;3x+4y−10...(i)3x + 4y - 10...(i) 2x−2y=2....(ii)2x - 2y = 2....(ii)

Solve the following systems equations by eliminations method;2x+3y=462x + 3y = 463x+5y=74 3x + 5y = 74

Solve the following systems equations by eliminations method;x+y=5x + y = 52x−3y=42x - 3y = 4

The sum of two numbers is 1616 and the sum of their reciprocals is 1/3. 1 / 3 . Find the numbers.

Solve the following equations by elimination method1x−1y=1\dfrac{1}{x} - \dfrac{1}{y} = 11x+1y=7.x≠0.y≠0\dfrac{1}{x} + \dfrac{1}{y} = 7.x \neq 0.y \neq 0

Two positive numbers differ by 3 and their product isFind the numbers.

Solve the following equations bu elimination method:99x+101y=49999x + 101y = 499101x+99y=501101x + 99y = 501

Solve the following equations bu elimination method:29x−23y=11029x - 23y = 11023x−29y=9823x - 29y = 98

Solve the following systems of equations by elimination method:x6+y15=4,x3−y12=194\dfrac{x}{6} + \dfrac{y}{15}= 4 , \dfrac{x}{3} - \dfrac{y}{12} =\dfrac{19}{4}

Solve the following system of equations by elimination method:x+6y=6,3x−8y=5x + \dfrac{6}{y} = 6 ,3x - \dfrac{8}{y} = 5

Solve the following system of equations by elimination method:2x+5y=12x + 5y = 12x+3y=32x + 3y = 3

Solve the following equations bu elimination method:217x+131y=913217x + 131y = 913And 131x+217y=827131x + 217y = 827

Solve the following systems of equations by elimination method:3x−8y=53x - \dfrac{8}{y} = 5x−y3=3 x - \dfrac{y}{3} = 3

Solve the following equations by elimination method4x+7y=11,3x−5y=−2 (x=≠0.y≠0)\dfrac{4}{x} + \dfrac{7}{y} = 11, \dfrac{3}{x} - \dfrac{5}{y} = -2\ \ (x = \neq 0.y \neq 0)

Two numbers differ by 2 and their product is 360. 360 . Find the numbers.

Two numbers differ by 4 and their product is 192 . Find the numbers.

Solve the following equations by elimination methodx=y=2xyx = y = 2xy , x−y=6xyx - y = 6xy

Solve the following equations by elimination method5x+y=19xy,7x−2y=8xy5x + y = 19xy , 7x - 2y = 8xy

Solve the following equations by elimination method3ax−2by+5=0,ax−3by−2=0(x=≠0.y≠0)\dfrac{3a}{x} - \dfrac{2b}{y} + 5 = 0 , \dfrac{a}{x} - \dfrac{3b}{y} - 2 = 0(x = \neq 0.y \neq 0)

Solve the following equations by elimination methodx+y=7xyx + y = 7xy , 2x−3y=−xy2x - 3y = -xy

A number consisting of two digits is seven times the sum of its digits. When 27 is subtracted from the number, the digits are reversed. Find the number.

Solve the following equations by elimination method2x+5yxy=6,4x−5yxy=−3(wherex=≠0andy≠0)\dfrac{2x + 5y}{xy} = 6 , \dfrac{4x - 5y}{xy} = -3(where x = \neq 0 and y \neq 0)

Solve the following equations by elimination method2x+3y=13\dfrac{2}{x} + \dfrac{3}{y} = 135x−4y=2.x≠0.y≠0\dfrac{5}{x} - \dfrac{4}{y} = 2.x \neq 0.y \neq 0

The age of the father is 3 years more than 3 times the son's age. 3 years here, the age of the father will be 10 years more than twice the age of the son. Find their present ages.

For which value of a, the following system of linear equations has no solutions :ax+3y=a−2,12x+ay=aax + 3y = a - 2, 12x + ay = a

Ten years hence, a man's age will be twice the age of his son. Ten years ago, the man was four times as old as his son. Find their present ages

Form the pairs of linear equations for the following problems and find their solutions by elimination method:Aftab tells his daughter,"seven years ago, I was seven times as old as you were then, Also, three years from now I shall be three times as old as you will be.'Find their present ages.

The sum of the digits of a two-digit number is 15. 15 . The number obtained by interchanging the digits exceeds the given number by 9. 9 . Find the number.

Check whether the following equations are consistent or inconsistent, solve them graphically.5x−3y=115x - 3y = 11 −10x+6y=−22 -10x + 6 y = -22

The sum of two digit number and the number obtained by reversing the digit is 66 . If the digit of the number differ by 2 . Find the number . How many such numbers are there?

Form the pairs of linear equations for the following problems and find their solutions by elimination method:Five years ago, Nuri was thrice as old as sonu. Ten years later, Nuri will be twice as old as sonu. How old are Nuri and Sonu?

Form the pairs of linear equations for the following problems and find their solutions by elimination method:The difference between two numbers is 26 and one number is three times the other find them

Five years ago, A was thrice as old as B and ten years later, A shall be twice as old as B. What are the present ages of A and B?

Solve each of the following pairs of equation by reducing then to a pair of linear equations.x+yxy=2,x−yxy=6\dfrac{x+y}{xy}=2,\dfrac{x-y}{xy}=6

Check whether the following equations are consistent or inconsistent , solve them graphically.2x -2y -2 = 0 and 4x - 4y -5 =0

Solve each of the following pairs of equation by reducing then to a pair of linear equations.13x+y+13x−y=34,12(3x+y)−12(3x−y)=−18\dfrac{1}{3x+y}+\dfrac{1}{3x-y}=\dfrac{3}{4},\dfrac{1}{2(3x+y)}-\dfrac{1}{2(3x-y)}=\dfrac{-1}{8}

Check whether the following equations are consistent or inconsistent , solve them graphically.2x + y - 6 = 0 and 4x - 2y -4 = 0

For solving each pair of equation, in this exercise, use the method of elimination by equating coefficients:x−y6=2(4−x) \frac{x-y}{6}=2(4-x) 2x+y=3(x−4) 2x+y=3(x-4)

For solving each pair of equation, in this exercise, use the method of elimination by equating coefficients:15(x−2)=14(1−y) \frac{1}{5}(x-2)=\frac{1}{4}(1-y) 26x+3y=−4 26x+3y=-4

For solving each pair of equation, in this exercise, use the method of elimination by equating coefficients:3−(x−5)=4+2 3-(x-5)=4+2 2(x+y)=4−3y 2(x+y)=4-3y

Solve each of the following pairs of equation by reducing then to a pair of linear equations.10x+y+2x−y=4,15x+y−5x−y=−2\dfrac{10}{x+y}+\dfrac{2}{x-y}=4,\dfrac{15}{x+y}-\dfrac{5}{x-y}=-2

For solving each pair of equation, in this exercise, use the method of elimination by equating coefficients:5y2−x3=8 \frac{5y}{2}-\frac{x}{3}=8 y2+5x3=12 \frac{y}{2}+\frac{5x}{3}=12

Check whether the following equations are consistent or inconsistent , solve them graphically.43x+2y=8,2x+3y=12 \dfrac{4}{3} x + 2y = 8 , 2x + 3y = 12

Solve the following simultaneous equation graphically.x−3y=1x - 3y = 1; 3x−2y+4=03x - 2y + 4 =0

The sum of digits of a two digit number is 11. 11 .If the digit at ten's place is increased by 5 5 and the digit at unit' place is decreased by 5 5 , the digits of the number are found to be reversed.Find the original number.

In the figure given above, ABCD is a parallelogram. Find the value of x and y.

Solve using the method of elimination by equating coefficients:2x−3y−3=0 2x-3y-3=0 2x3+4y+12=0 \dfrac{2x}{3}+4y+\dfrac{1}{2}=0

From the pair of leaner equation in the following problems , and find their solution graphically .5 pencils and 7 pens together cost Rs , 50 whereas 7 pencils and 5 pens together cost Rs . 46 .Find the cost of a pencils and on one pen .

A A and B B both have some pencils . if A A gives 10 10 pencils to B B , then B B will have twice as many as A A . And if B B gives 10 10 pencils to A A , then they will have the same number of pencils . how many pencils does each have ?

The graph of 3x+2y=63x + 2y = 6 meets the x=axisx=axis at point PP and the y−axisy-axis at point QQ. Use the graphical method to find the co-ordinates of points PP and QQ.

Solve: 11(x−5)+10(y−2)+54=0 11(x-5)+10(y-2)+54=0 7(2x−1)+9(3y−1)=25 7(2x-1)+9(3y-1)=25

Given the linear equation 2x+3y−8=0 2x + 3 y - 8 = 0 write linear equation in two variable such that the geometrical representation of the pair so formed is :Intersecting lines

Solve the following pair of linear equation by the substituting method .0.2x+0.3y=1.3 0 . 2x + 0 . 3y = 1.3 0.4x+0.5y=2.3 0.4 x + 0.5y = 2.3

Solve the following pair of linear equation by the substituting method .3x2−5y3=−2 \dfrac{3x}{2} - \dfrac{5y}{3} = -2 x3+y2=132 \dfrac{x}{3} + \dfrac{y}{2} = \dfrac{13}{2}

Form the pair of linear equation for the following problems and find their solution by substituting method .The difference between two number is 26 26 and one number is three times the other. Find them.

Form the pair of linear equation for the following problems and find their solution by substituting method .The larger of two supplementary angles axceeds the smaller by 18 18 degrees . Find them .

Solve the following pair of linear equation by the substituting method .s−t=3 s - t = 3 s3+t2=6 \dfrac{s}{3} + \dfrac{t}{2} = 6

Solve the following pair of linear equation by the substituting method .3x−y=3 3x - y = 3 9x−3y=9 9x - 3y = 9

Solve the following pair of linear equation by the substituting method .x+y=14 x + y = 14 x−y=14 x - y = 14

Form the pair of linear equation for the following problems and find their solution by substituting method .The coach of a cricket them buys 7 7 bats and 6 6 balls for Rs 3800 3800 . Later , she buys 3 3 bats and 5 5 balls for Rs 1750 1750 Find the cost of each bat and each ball.

Solve the following pair of linear equation by the substituting method .√2x+√3y=0 \sqrt{2x} + \sqrt{ 3y} = 0 √3x−√8y=0 \sqrt{3x} - \sqrt{8 y} = 0

Solve the following equations graphically: $3x-2y = 4$ , $5x-2y = 0$

Solve the following pairs of linear equations by substition method;
$3x - y = 3$
$9x - 3y = 9$

Solve the following pair of linear equations by substitution method:
$x + y = 2m$
$mx - ny = m^2+n^2$

Solve the following pair of linear equations by substitution method:
$\dfrac{x}{2} + y = 0.8$
$x + \dfrac{y}{2} = \dfrac{7}{10}$

Solve the following pair of linear equations by substitution method:
$\dfrac{x}{a} +\dfrac{y}{b} = 2 , a\neq 0,b \neq 0$
$ax - by = a^2 - b^2$

Solve graphically the following system of linear equations if it has unique solution :
$3x + y = 2$
$6x + 2y = 1$

Find graphically the co-ordinates of the vertices of the triangle formed by the lines $y = 0, y = x$ and $2x + 3y= 10$ . Hence find the area of the triangle formed by these lines.

Solve the following pair of linear equations by substitution method:
0.5x - 0.8y = 3.4
0.6x + 0.3y = 0.3

Solve the following pair of linear equations by substitution method:
$S - t = 3$
$\dfrac{s}{3} + \dfrac{t}{2} = 6$

Using the same axes of co-ordinates and the same unit, solve graphically.
$x+y = 0, 3x 2y = 10$

Solve the following pair of linear equations by substitution method:
$7x - 15y = 2$
$x + 2y = 3$

Solve the following systems equations by eliminations method;
$3x - 5y - 4 = 0...(i)$
$9x = 2y + 7....(ii)$

Solve the following systems equations by eliminations method;
$8x + 5y = 9$
$3x + 2y = 4$

Solve the following systems equations by eliminations method;
$0.4x + 1.5y = 6.5$
$0.3x - 0.2y = 0.9$

Solve the following systems equations by eliminations method;
$\sqrt{2}x - \sqrt{3}y = 0$
$\sqrt{5}x + \sqrt{.2}y = 0$

Solve the following systems equations by eliminations method;
$2x + 3y = 8$
$4x + 6y =7$

Solve the following pair of linear equations by substitution method:
$\dfrac{bx}{a} + \dfrac{ay}{b} = a^2 + b^2$
$x + y = 2ab$

Solve graphically the following system of linear equations if is has unique solution:
$2x - 6y + 10 = 0$
$3x - 9y + 15 =0$

Solve the following systems equations by eliminations method;
$3x + 4y - 10...(i)$
$2x - 2y = 2....(ii)$

Solve the following systems equations by eliminations method;
$2x + 3y = 46$
$3x + 5y = 74$

Solve the following systems equations by eliminations method;
$x + y = 5$
$2x - 3y = 4$

The sum of two numbers is $16$ and the sum of their reciprocals is $1 / 3 .$ Find the numbers.

Solve the following equations by elimination method
$\dfrac{1}{x} - \dfrac{1}{y} = 1$
$\dfrac{1}{x} + \dfrac{1}{y} = 7.x \neq 0.y \neq 0$

Solve the following equations bu elimination method:
$99x + 101y = 499$
$101x + 99y = 501$

Solve the following equations bu elimination method:
$29x - 23y = 110$
$23x - 29y = 98$

Solve the following systems of equations by elimination method:
$\dfrac{x}{6} + \dfrac{y}{15}= 4 , \dfrac{x}{3} - \dfrac{y}{12} =\dfrac{19}{4}$

Solve the following system of equations by elimination method:
$x + \dfrac{6}{y} = 6 ,3x - \dfrac{8}{y} = 5$

Solve the following system of equations by elimination method:
$2x + 5y = 1$
$2x + 3y = 3$

Solve the following equations bu elimination method:
$217x + 131y = 913$
And $131x + 217y = 827$

Solve the following systems of equations by elimination method:
$3x - \dfrac{8}{y} = 5$
$x - \dfrac{y}{3} = 3$

Solve the following equations by elimination method
$\dfrac{4}{x} + \dfrac{7}{y} = 11, \dfrac{3}{x} - \dfrac{5}{y} = -2\ \ (x = \neq 0.y \neq 0)$

Two numbers differ by 2 and their product is $360 .$ Find the numbers.

Solve the following equations by elimination method
$x = y = 2xy$ ,
$x - y = 6xy$

Solve the following equations by elimination method
$5x + y = 19xy , 7x - 2y = 8xy$

Solve the following equations by elimination method
$\dfrac{3a}{x} - \dfrac{2b}{y} + 5 = 0 , \dfrac{a}{x} - \dfrac{3b}{y} - 2 = 0(x = \neq 0.y \neq 0)$

Solve the following equations by elimination method
$x + y = 7xy$ ,
$2x - 3y = -xy$

Solve the following equations by elimination method
$\dfrac{2x + 5y}{xy} = 6 , \dfrac{4x - 5y}{xy} = -3(where x = \neq 0 and y \neq 0)$

Solve the following equations by elimination method
$\dfrac{2}{x} + \dfrac{3}{y} = 13$
$\dfrac{5}{x} - \dfrac{4}{y} = 2.x \neq 0.y \neq 0$

For which value of a, the following system of linear equations has no solutions :
$ax + 3y = a - 2, 12x + ay = a$

Form the pairs of linear equations for the following problems and find their solutions by elimination method:
Aftab tells his daughter,"seven years ago, I was seven times as old as you were then, Also, three years from now I shall be three times as old as you will be.'Find their present ages.

The sum of the digits of a two-digit number is $15 .$ The number obtained by interchanging the digits exceeds the given number by $9 .$ Find the number.

Check whether the following equations are consistent or inconsistent, solve them graphically.
$5x - 3y = 11$
$-10x + 6 y = -22$

Form the pairs of linear equations for the following problems and find their solutions by elimination method:
Five years ago, Nuri was thrice as old as sonu. Ten years later, Nuri will be twice as old as sonu. How old are Nuri and Sonu?

Form the pairs of linear equations for the following problems and find their solutions by elimination method:
The difference between two numbers is 26 and one number is three times the other find them

Solve each of the following pairs of equation by reducing then to a pair of linear equations.
$\dfrac{x+y}{xy}=2,\dfrac{x-y}{xy}=6$

Check whether the following equations are consistent or inconsistent , solve them graphically.
2x -2y -2 = 0 and 4x - 4y -5 =0

Solve each of the following pairs of equation by reducing then to a pair of linear equations.
$\dfrac{1}{3x+y}+\dfrac{1}{3x-y}=\dfrac{3}{4},\dfrac{1}{2(3x+y)}-\dfrac{1}{2(3x-y)}=\dfrac{-1}{8}$

Check whether the following equations are consistent or inconsistent , solve them graphically.
2x + y - 6 = 0 and 4x - 2y -4 = 0

For solving each pair of equation, in this exercise, use the method of elimination by equating coefficients:

$\frac{x-y}{6}=2(4-x)$
$2x+y=3(x-4)$

For solving each pair of equation, in this exercise, use the method of elimination by equating coefficients:

$\frac{1}{5}(x-2)=\frac{1}{4}(1-y)$
$26x+3y=-4$

For solving each pair of equation, in this exercise, use the method of elimination by equating coefficients:

$3-(x-5)=4+2$
$2(x+y)=4-3y$

Solve each of the following pairs of equation by reducing then to a pair of linear equations.
$\dfrac{10}{x+y}+\dfrac{2}{x-y}=4,\dfrac{15}{x+y}-\dfrac{5}{x-y}=-2$

For solving each pair of equation, in this exercise, use the method of elimination by equating coefficients:

$\frac{5y}{2}-\frac{x}{3}=8$
$\frac{y}{2}+\frac{5x}{3}=12$

Check whether the following equations are consistent or inconsistent , solve them graphically.
$\dfrac{4}{3} x + 2y = 8 , 2x + 3y = 12$

Solve the following simultaneous equation graphically.
$x - 3y = 1$ ; $3x - 2y + 4 =0$

The sum of digits of a two digit number is $11 .$ If the digit at ten's place is increased by $5$ and the digit at unit' place is decreased by $5$ , the digits of the number are found to be reversed.Find the original number.

Solve using the method of elimination by equating coefficients:

$2x-3y-3=0$

$\dfrac{2x}{3}+4y+\dfrac{1}{2}=0$

From the pair of leaner equation in the following problems , and find their solution graphically .
5 pencils and 7 pens together cost Rs , 50 whereas 7 pencils and 5 pens together cost Rs . 46 .Find the cost of a pencils and on one pen .

$A$ and $B$ both have some pencils . if $A$ gives $10$ pencils to $B$ , then $B$ will have twice as many as $A$ . And if $B$ gives $10$ pencils to $A$ , then they will have the same number of pencils . how many pencils does each have ?

The graph of $3x + 2y = 6$ meets the $x=axis$ at point $P$ and the $y-axis$ at point $Q$ . Use the graphical method to find the co-ordinates of points $P$ and $Q$ .

Solve:
$11(x-5)+10(y-2)+54=0$
$7(2x-1)+9(3y-1)=25$

Given the linear equation $2x + 3 y - 8 = 0$ write linear equation in two variable such that the geometrical representation of the pair so formed is :
Intersecting lines

Solve the following pair of linear equation by the substituting method .
$0 . 2x + 0 . 3y = 1.3$
$0.4 x + 0.5y = 2.3$

Solve the following pair of linear equation by the substituting method .
$\dfrac{3x}{2} - \dfrac{5y}{3} = -2$
$\dfrac{x}{3} + \dfrac{y}{2} = \dfrac{13}{2}$

Form the pair of linear equation for the following problems and find their solution by substituting method .
The difference between two number is $26$ and one number is three times the other. Find them.

Form the pair of linear equation for the following problems and find their solution by substituting method .
The larger of two supplementary angles axceeds the smaller by $18$ degrees . Find them .

Solve the following pair of linear equation by the substituting method .
$s - t = 3$
$\dfrac{s}{3} + \dfrac{t}{2} = 6$

Solve the following pair of linear equation by the substituting method .
$3x - y = 3$
$9x - 3y = 9$

Solve the following pair of linear equation by the substituting method .
$x + y = 14$
$x - y = 14$

Form the pair of linear equation for the following problems and find their solution by substituting method .
The coach of a cricket them buys $7$ bats and $6$ balls for Rs $3800$ . Later , she buys $3$ bats and $5$ balls for Rs $1750$ Find the cost of each bat and each ball.

Solve the following pair of linear equation by the substituting method .
$\sqrt{2x} + \sqrt{ 3y} = 0$
$\sqrt{3x} - \sqrt{8 y} = 0$

Solve the following pair of linear equations graphically and write nature of solution.
$2x - y = 4; x + y = -1$

Solve the following pair of linear equations graphically and find the coordinates of the triangle so formed with the y-axis and the lines.
$4x - 5y = 20,$
$3x + 5y = 15$

Solve the following pair of linear equations, graphically and find the coordinates of that points where lines represented by these cuts y-axis.
$2x - 5y + 4 = 0; 2x + y - 8 = 0$

Solve the following pair of linear equations graphically and find the coordinates of that points where lines are represented by these cuts y-axis.
$3x + 2 = 12; 5x - 2y = 4$

Solve the following systems of simultaneous linear equations by the method of elimination by equating the coefficient.
$\dfrac{x}{7} + \dfrac{y}{3} = 5$
$\dfrac{x}{2} - \dfrac{y}{9} = 6$

Solve graphically:
$3x - 4y = 1; -2x + \dfrac{8}{3}y = 5$

Solve the following systems of simultaneous linear equations by the method of elimination by equating the coefficient.
$2x + y = 13$
$5x - 3y = 16$

Solve the following systems of simultaneous linear equations by the method of substitution.
$8x + 5y = 9$
$3x + 2y = 4$

Solve the following systems of simultaneous linear equations by the method of substitution.
$3x + 2y = 11$
$2x + 3y = 4$