MCQ Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 9

Solve graphically the pair of equations

$x+3y=6$ , and

$3x-5y=18$ . Hence, find the value of

$K$ if

$7x+3y=K$

0%
$x=-3, y=1, K=13$
0%
$x=-8, y=5, K=8$
0%
$x=1, y=2, K=29$
0%
$x=6, y=0, K=42$

Explanation

From the graph the point of intersection is

$(x,y)=(6,0)$ .

Substituting

$(x,y)=(6,0)$ in

$7x+3y=K$ , we get

$7\times 6+3\times 0=K$

i.e

$K=42$

Ans- Option D.

$2x + y = 23$ and

$4x - y = 19$ , find the values of

$5y - 2x$ and

$\dfrac{y}{x} - 2$ .

0%
$36, -\dfrac13$
0%
$31, -\dfrac57$
0%
$38, \dfrac67$
0%
None of these

Explanation

Given System of Equations are:

$2x+y = 23$ ...

$(1)$

$4x-y = 19$ ...

$(2)$
Now adding both the equation will have:

$6x = 42$

$x = 7$
Now substituting

$x = 7$ in in equation

$(1)$

$\Rightarrow 2\left ( 7 \right ) + y = 23$

$\Rightarrow 14 + y = 23$

$\Rightarrow y = 9$
Now substituting

$x = 7$ and

$y = 9$ in

$5y - 2x$

$\Rightarrow 5\left ( 9 \right ) - 2\left ( 7 \right )$

$\Rightarrow 45-14$

$\Rightarrow 31$
Now Substituting Now substituting

$x = 7$ and

$y = 9$ in

$\frac{y}{x}-2$

$\Rightarrow \dfrac{9}{7}-2$

$\Rightarrow \dfrac{9-14}{7}$

$\Rightarrow \dfrac{-5}{7}$

Determine by drawing graphs whether the following pair of equations has a unique solution or not:

$2x - 3y = 6, 4x - 6y = 9$ . If yes, find the solution also

0%
No
0%
Yes, $x=3$ , $y=-2$
0%
Ambiguous
0%
Data insufficient

Find the value of

$k$ for which the given system of equations has infinite number of solutions.

$5x + 2y = 2k$ and

$2(k+ 1)x + ky = (3k+ 4)$

0%
4
0%
7
0%
3
0%
6

Explanation

Comparing

$5x+2y=2k$ with

$a_1x+b_1y+c_1=0$ , we get,

$a_1=5, b_1=2,c_1=-2k$

Comparing

$2(k+1)x+ky=3k+4$ with

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

$a_2=2(k+1), b_2=k, c_2=-(3k+4)$

Now,

Given: the equations has infinite no of solutions.

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

$\dfrac{5}{2(k+1)}=\dfrac{2}{k}=\dfrac{-2k}{-(3k+4)}$

When,

$\dfrac{5}{2(k+1)}=\dfrac{2}{k}$

$\implies 5k=4(k+1)$

$\implies 5k-4k=4$

$\implies k=4$

and

$\dfrac{2}{k}=\dfrac{-2k}{-(3k+4)}$

$\implies 2(3k+4)=2k^2$

$\implies 3k+4=k^2$

$\implies k^2-3k-4=0$

$\implies k^2-4k+k-4=0$

$\implies k(k-4)+(k-4)=0$

$\implies (k-4)(k+1)=0$

Either,

$k=4\space or\space k=-1$

$k=4$ is the value that satisfies both the conditions.

So,

$Option \ A \ is\ correct$

Solve:

$\frac {4}{9}x+\frac {1}{3}y=1, 5x+2y=13$

0%
$x=3, y=-5$
0%
$x=3, y=-6$
0%
$x=3, y=-1$
0%
$x=1, y=0$

Explanation

Given system of equations are:

$\frac{4}{9}x + \frac{1}{3}y = 1$ ... (1)

$5x+2y = 13$ ... (2)
From equation 1 will get:

$4x + 3y = 9$
Now subtracting eq1 and eq2

$4x +3y = 9$

$5x +2y = 13$
First multiply eq1 from 2 and eq2 from 2 will get:

$8x + 6y = 18$

$15x+6y = 39$
now will subtract the equation

$\Rightarrow 8x +6y = 18$

$-15x -6y = -39$

$\Rightarrow -7x = -21$

$\Rightarrow x = 3$
now substituting x = 3 in eq 2

$\Rightarrow 15\left ( 3 \right )+ 6y = 39$

$\Rightarrow 45 + 6y = 39$

$\Rightarrow 6y = 39-45$

$\Rightarrow 6y = -6$

$\Rightarrow y = -1$

$\Rightarrow x = 3 , y = -1$

Determine how many solution exist for the following pair of equations

$8x +5y = 9, 16x +10y = 27$

0%
Both equations form coincident line hence they have no solutions.
0%
Both equations form parallel lines hence they have no solutions.
0%
Both equations form intersecting lines hence they have no solutions.
0%
None of these

Explanation

The following pair of equations can be written as

$8x+5y-9=0 \ ; 16x+10y-27=0;$

Here

$\displaystyle \frac{a_1}{a_2}=\frac{8}{16}=\frac{1}{2}$

$\dfrac{b_1}{b_2}=\dfrac{5}{10}=\dfrac{1}{2}$

$\dfrac{c_1}{c_2}=\dfrac{-9}{-27}=\dfrac{1}{3}$ ,

We can see

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

Hence, the given pair of the linear equation is inconsistent and has no solution.

For a graphical solution,

$8x+5y=9$

$y=\dfrac{9-8x}{5}$

$16x+10y=27$

$y=\dfrac{27-16x}{10}$

Now, we can make the table and make the graph as shown in the solution.

Determine whether the system of linear equations

$2x + 3y - 5 = 0, 6x + 9y - 15 = 0$ has a unique solution, no solutions, or an infinite number of solutions.

0%
Infinite number of solutions
0%
No solutions
0%
Unique solution
0%
Cannot be determined

Explanation

The equations are

$\Rightarrow 2x+3y-5=0$
here,

$a_{1}=2,b_{1}=3,c_{1}=-5$

$\Rightarrow 6x+9y-15=0$
here,

$a_{2}=6,b_{2}=9,c_{2}=-15$

$\Rightarrow \displaystyle \frac{a_{1}}{a_{2}}=\frac{2}{6}= \frac{1}{3}$

$\Rightarrow \displaystyle \frac{b_{1}}{b_{2}}=\frac{3}{9}=\frac{1}{3}$

$\Rightarrow \displaystyle \frac{c_{1}}{c_{2}}=\frac{-5}{-15}=\frac{1}{3}$

$\Rightarrow \displaystyle \frac{a_{1}}{b_{1}}=\frac{a_{2}}{b_{2}}=\frac{c_{1}}{c_{2}}$

Hence, the given system of equations has an infinite number of solutions.

Solve the following equations by the substitution method

$\dfrac {1}{2}(9x+10y)=23, \dfrac {5x}{4}-2y=3$

0%
$x = 4, y =1$
0%
$x = 2, y =5$
0%
$x = 1, y =-1$
0%
$x = 7, y =-3$

Explanation

The equations are

$\Rightarrow \frac{1}{2}\left ( 9x+10y \right )=23\Rightarrow 9x+10y=46....eq\left ( 1 \right )$

$\Rightarrow x=\frac{46-10y}{9}$

$\Rightarrow \frac{5x}{4}-2y=3\Rightarrow 5x-8y=12....eq\left ( 2 \right )$
Substitute the value of x in eq (2 )

$\Rightarrow \frac{5\left (46-10y \right )}{9}-8y=12$

$\Rightarrow 230-50y-72y=108\Rightarrow -122y=-122\Rightarrow y=1$
Substitute the value of y in eq (2 )

$\Rightarrow 5x-8\left (1 \right )=12\Rightarrow 5x=20\Rightarrow x=4$

$\Rightarrow x=4,y=1$

Solve the following equations by the substitution method

$0.04 x + 0.02y = 5, 0.5x - 0.4y = 30$

0%
$x=27, y=61$
0%
$x=100, y=50$
0%
$x=200, y=39$
0%
$x=54, y=122$

Explanation

The equations are

$\Rightarrow .04x+.02y=5....eq\left ( 1 \right )$

$\Rightarrow x=\frac{5-.02y}{.04}$

$\Rightarrow .5x-.4y=30....eq\left (2 \right )$
Substitute the value of x in eq (2)

$\Rightarrow \frac{.5\left (5-.02y \right )}{.04}-.4y=30$

$\Rightarrow 2.5-.01y-.016y=1.2\Rightarrow -.026y=-1.3\Rightarrow y=50$
Substitute the value of y in eq (2)

$\Rightarrow .5x-.4\left (50 \right )=30\Rightarrow .5x=50\Rightarrow x=100$

$\Rightarrow x=100,y=50$

Solve the following equations by the substitution method

$x = 3y - 19, y = 3x - 23$

0%
$x = 5, y = 7$
0%
$x = 11, y = 10$
0%
$x = 13, y = 7$
0%
$x = 3, y = 11$

Explanation

The equations are

$\Rightarrow x=3y-19....eq\left ( 1 \right )$

$\Rightarrow y=3x-23....eq\left ( 2 \right )$
Substitute the value of x in eq (2 )

$\Rightarrow y=3\left (3y-19 \right )-23$

$\Rightarrow y=9y-57-23\Rightarrow -8y=-80\Rightarrow y=10$
Substitute the value of y in eq (1 )

$\Rightarrow x=3\left (10 \right )-19\Rightarrow x=30-19\Rightarrow x=11$

$\Rightarrow x=11,y=10$

Solve the following pairs of linear equations by elimination method:

$78x + 91y = 39$ and

$65x + 117y = 42$

0%
$x = \dfrac{3}{13}, y = \dfrac{3}{13}$
0%
$x = \dfrac{1}{11}, y = \dfrac{2}{11}$
0%
$x = \dfrac{5}{17}, y = \dfrac{4}{17}$
0%
Cannot be determined

Explanation

Given system of equations are:

$78x+91y = 39 ...(1)$

$65x+117y =42 ...(2)$
Now using elimination method will first eliminate

$x$ .
now multiplying eq

$(1)$ by

$65$ and eq

$(2)$ by

$-78$ .

$\Rightarrow \left ( 65 \right )78x + \left ( 65 \right )91y = \left ( 65 \right )39$

$\Rightarrow \left ( -78 \right )65x + \left ( 78 \right )117y = \left ( 65 \right )42$

$\Rightarrow 5070x + 5915y = 2535 ...(3)$

$\Rightarrow -5070x - 9126y = -3276 ...(4)$
From equation

$(3)$ and

$(4)$ , we get

$-3211y = -741$

$\Rightarrow y = \dfrac{-741}{-3211}$

$y = \dfrac{3}{13}$
Now substituting

$y = \dfrac{3}{13}$ in eq

$(1)$

$\Rightarrow 78x + 91\left ( \dfrac{3}{13} \right )= 39$

$\Rightarrow 1014x +273 = 507$

$\Rightarrow 1014x = 507 - 273$

$\Rightarrow 1014x = 234$

$\Rightarrow x = \dfrac{234}{1014}$

$\Rightarrow x = \dfrac{3}{13}$

$\Rightarrow x = \dfrac{3}{13} , y = \dfrac{3}{13}$

Solve the following pair of linear equations by elimination method

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

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

0%
$x = 1, y =7$
0%
$x = 2, y =- 3$
0%
$x = 3, y =- 3$
0%
$x = 5, y =-7$

Explanation

Given system of equations are:

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

$x - \frac{y}{3} = 3$ ................(2)
From eq 1 will have

$\Rightarrow 3x+4y = - 6 .$ ........(3)
From eq 2 will have

$\Rightarrow 3x-y = 9$ ...........(4)

Now using elimination method frist we will eliminate y :
now multiplying eq 4 from 4 will have

$\Rightarrow 12x-4y = 36$
Now from eq 3 and 4 will get x:

$3x+4y = -6$

$12x-4y = 36$

$\Rightarrow 15x = 30$

$\Rightarrow x = 2$
Now substituting

$x = 2$ in eq 2 will get y

$\Rightarrow \left ( 2 \right )3 - y = 9$

$\Rightarrow 6 - y = 9$

$\Rightarrow -y = 9-6$

$\Rightarrow y = -3$

$\Rightarrow x = 2 , y = -3$

Solve graphically the following pair of equations:

$x -y = 1, 2x + y= 8$ . Shade the area bounded by these lines and the y-axis

0%
$x = 1, y = -1$
0%
$x = 2, y = 0$
0%
$x = 4, y = -1$
0%
$x = 3, y = 2$

Solve the following pair of equations graphically :

$x + y = 4, 3x - 2y =- 3$
Shade the region bounded by the lines representing the above equations and x-axis

0%
$x = 3, y = 2$
0%
$x = 1, y = 3$
0%
$x = 8, y = 2$
0%
$x = 9, y = 2$

Explanation

Equation (i) is

$x+y=4$

i.e.

$y=4-x$ and

$3x-2y=-3$

i.e

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

The point of intersection from the graphs is

$(x,y)=(1,3)$

The required area is a triangle

Ans- Option B.

Solve the following equations by the substitution method

$11x-8y=27, 3x+5y=-7$

0%
$x=0, y=1$
0%
$x=0, y=-5$
0%
$x=2, y=3$
0%
$x=1, y=-2$

Explanation

The equations are

$\Rightarrow 11x-8y=27....eq\left ( 1 \right )$

$\Rightarrow x=\frac{27+8}{11}$

$\Rightarrow 3x+5y=-7....eq\left ( 2 \right )$
Substitute the value of x in eq(2 )

$\Rightarrow \frac{3\left (27+8y \right )}{11}+5y=-7$

$\Rightarrow 81+24y+55y=-77\Rightarrow 79y=-158\Rightarrow y=-2$
Substitute the value of y in eq (2 )

$\Rightarrow 3x+5\left (-2 \right )=-7\Rightarrow 3x=3\Rightarrow x=1$

$\Rightarrow x=1,y=-2$

Solve the following equations by the substitution method

$\dfrac {x+11}{7}+2y=10, 3x=8+\dfrac {y+7}{11}$

0%
$x = 1, y = -2$
0%
$x = 8, y = -7$
0%
$x = 9, y = 2$
0%
$x = 3, y = 4$

Explanation

The equations are

$\Rightarrow \dfrac{x+11}{7}+2y=10\Rightarrow x+11+14y=70$

$\Rightarrow x+14y=59 ....eq\left ( 1 \right )$

$\Rightarrow x=59-14y$

$\Rightarrow 3x=8+\dfrac{y+7}{11}\Rightarrow 33x=88+y+7$

$\Rightarrow 33x-y=95....eq\left ( 2 \right )$
Substitute the value of x in eq (2 )

$\Rightarrow 33\left (59-14y \right )-y=95$

$\Rightarrow 1947-462y-y=95\Rightarrow 463y=1853\Rightarrow y=4$

$\Rightarrow x=59-14y\Rightarrow x=59-14\left (4 \right ) \Rightarrow x=3$

$\Rightarrow x=3,y=4$

Find the values of x and y in the following rectangle

0%
$x=7, y=-8$
0%
$x=1, y=-5$
0%
$x=2, y=0$
0%
$x=1, y=4$

Explanation

${\textbf{Step-1: Use rectangle property.}}$

${\text{As we know that opposite sides of a rectangle are equal.}}$

${\text{According to the question equations are}}$

$\Rightarrow x+3y=13$

$...(1)$

$\Rightarrow x=13-3y$

$\Rightarrow 3x+y=7$

$...(2)$

${\textbf{Step-2: Solve the equations to find the values of x and y}.}$

${\text{Put the value of}}$

$x$

${\text{in equation (2).}}$

$\Rightarrow 3(13-3y)+y=7$

$\Rightarrow 39 -9y + y = 7$

$\Rightarrow -8y =-32$

$\Rightarrow y = 4$

${\text{Put the value of}}$

$y$

${\text{ in equation (1).}}$

$\Rightarrow x=13-3y$

$\Rightarrow x=13-3 \times 4$

$\Rightarrow x=13-12$

$\Rightarrow x=1$

${\textbf{Hence, the value of x = 1 and y = 4. So, option D is correct.}}$

Based on equations reducible to linear equations, Solve for x and y:

$6x + 5y = 8xy$ and

$8x + 3y = 7xy$

0%
$x = 2, y = 2$
0%
$x = 1, y = 2$
0%
$x = 3, y = 8$
0%
$x = 5, y = 7$

Explanation

The given equations are
$\Rightarrow 6x+5y=8xy....eq\left (1 \right )$

$\Rightarrow 8x+3y=7xy....eq\left (2 \right )$

$Divide \ both \ equations \ by \ \ xy$

$\Rightarrow \dfrac{6}{y}+\dfrac{5}{x}=8....eq\left ( 3 \right )$

$\Rightarrow \dfrac{8}{y}+\dfrac{3}{1x}=7....eq\left ( 4 \right )$

$Put\ \dfrac{1}{x}=u\ and\ \dfrac{1}{y}=v\ in\ eq\left (3 \right ) and \left ( 4 \right )$

$\Rightarrow 6v+5u=8....eq\left ( 5 \right )$

$\Rightarrow 8v+3u=7 ....eq\left ( 6\right )$

$Multiply \ eq \ \left (5 \right )\ by \ 3 \ and \ eq\left ( 6 \right ) \ by\ 5 \ and \ subtract \ both$

$\Rightarrow \left (18v+15u=24 \right )-\left ( 40v+15u=35 \right )$

$\Rightarrow -22v=-11\Rightarrow v=\dfrac{1}{2}$

$Put \ v=\dfrac{1}{2} \ in \ eq\left ( 5\right )$

$\Rightarrow 6\left ( \dfrac{1}{2} \right )+5u=8\Rightarrow u=1$

$Hence,\dfrac{1}{x}= u=1\Rightarrow x=1$

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

Based on equations reducible to linear equations
Solve for x and y:

$9 + 25xy = 53x$ and

$27 - 4xy = x$

0%
$x = 1, y = 4$
0%
$x = 2, y = 7$
0%
$x = 6, y = 2$
0%
$x = 3, y = 2$

Explanation

Given systems of equations are:
$9+25xy = 53x$
$27-4xy = x$
Now using reducible method will get
Let convert this equation in $\dfrac{1}{x}\ and\ \dfrac{1}{y}$ form
Will take $xy$ as $L.C.M$ as $xy$ is common
From eq will get
$\Rightarrow \dfrac{9}{xy} +25 = \dfrac{53}{y}$ ...3
$\dfrac{27}{xy} - 4 = \dfrac{1}{y}$ ...4
Now let $\dfrac{1}{xy} = v and \dfrac{1}{y} = u$
Now will get
$9v+25 = 53u$ and $27v-4 = u$
Now Rearranging both the equation
$53u-9v = 25 ...(5)$
$u-27v = -4 ....(6)$
Now multiplying eq5 by 3
$\Rightarrow 159u-27v = 75 ...(7)$
Now subtracting eq 7 and eq6
$\Rightarrow 159u-27v = 75$
$\Rightarrow -u +27v = 4$
$\Rightarrow 158u = 79$
$\Rightarrow u = \dfrac{1}{2}$
Now putting $u = \dfrac{1}{2}$ in eq 6
$\Rightarrow \dfrac{1}{2} -27\left ( 2 \right ) v= -4\left ( 2 \right )$
$\Rightarrow 1 - 54v = -8$
$\Rightarrow 54v = 9$
$\Rightarrow v = \dfrac{9}{54}$
$\Rightarrow v = \dfrac{1}{6}$
Now putting u and v values
$\dfrac{1}{y} = u$
$\Rightarrow \dfrac{1}{y} = \dfrac{1}{2}$
$\Rightarrow y = 2$
$\dfrac{1}{xy} = v$
$\Rightarrow \dfrac{1}{xy} = \dfrac{1}{6}$
$\Rightarrow \dfrac{1}{2x} = \dfrac{1}{6}$
$\Rightarrow x = \dfrac{6}{2}$
$\Rightarrow x = 3$
$\Rightarrow x = 3 , y = 2$

Based on equations reducible to linear equations
Solve for x and y:

$\dfrac {11}{2x}-\dfrac {9}{2y}=-\dfrac {23}{2}; \dfrac {3}{4x}+\dfrac {7}{15y}=\dfrac {23}{6}$

0%
$x = 1/2, y = 1/5$
0%
$x = 1/5, y = 1/9$
0%
$x = 1/7, y = 1/2$
0%
$x = 1/3, y = 1/4$

Explanation

The given equations are
$\Rightarrow \dfrac{11}{2x}-\dfrac{9}{2y}=-\dfrac{23}{2}....eq\left ( 1 \right )$

$\Rightarrow \dfrac{3}{4x}+\dfrac{7}{15y}=-\dfrac{23}{6}....eq\left ( 2 \right )$

Put $u=\dfrac{1}{x}$ and $v=\dfrac{1}{y}$ in $eq\left (1 \right )$ and $\left ( 2 \right )$

$\Rightarrow \dfrac{11}{2}u-\dfrac{9}{2}v=-\dfrac{23}{2}\Rightarrow 11u-9v=-23....eq\left ( 3 \right )$

$\Rightarrow \dfrac{3}{4}u+\dfrac{7}{15}v=\dfrac{23}{6}\Rightarrow 45u+28v=230 ....eq\left ( 4 \right )$

Multiply $eq \left (3 \right )$ by 28 and $eq\left ( 4 \right )$ by 9 and subtract both

$\Rightarrow \left ( 308u-252v=-644 \right )-\left ( 405u+252v=2070 \right )$

$\Rightarrow 713u=1426\Rightarrow u=2$

$Put\ u=2\ in\ eq\left ( 3 \right )$

$\Rightarrow 11\times2-9v=-23\Rightarrow v=5$

$Hence, \dfrac{1}{x}= u=2\Rightarrow x=\dfrac{1}{2}$
$\dfrac{1}{y}= v=5\Rightarrow y=\dfrac{1}{5}$

Based on equations reducible to linear equations
Solve for x and y

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

0%
$x = 0, y = 1$
0%
$x = 0, y = -7$
0%
$x = 2, y = 3$
0%
$x = 2, y = 7$

Explanation

The given equations are

$\Rightarrow \frac{2}{x}+\frac{3}{y}=2....eq\left ( 1 \right )$

$\Rightarrow \frac{1}{x}-\frac{1}{2y}=\frac{1}{3}....eq\left ( 2 \right )$

$Put \frac{1}{x}=u \ and \ \frac{1}{y}=v \ in\ eq\left ( 1 \right ) \ and \ \left (2 \right )$

$\Rightarrow 2u+2v=2....eq\left (3 \right )$

$\Rightarrow u-\frac{1}{2}v=\frac{1}{3}\Rightarrow 6u-3v=2.....eq\left (4 \right )$

$Substract \ eq\left ( 3 \right ) \ and \ eq\left ( 4 \right )$

$\Rightarrow 8u=4\Rightarrow u=\frac{1}{2}$

$put \ the\ value \ of \ u \ in \ eq\left ( 3 \right )$

$\Rightarrow 2\times \frac{1}{2}+3v=2\Rightarrow v =\frac{1}{3}$

$Hence, \frac{1}{x}=u=\frac{1}{2}\Rightarrow x=2$

$\frac{1}{y}=v=\frac{1}{2}\Rightarrow y=3$

On the same axes, draw the graph of each of the following equations:

$2y-x = 8, 5y-x = 14, y- 2x = 1$ . Hence, obtain the vertices of the triangle so formed.

0%
$(1, 2), (-1, 2), (1, 6)$
0%
$(2, 5), (-4, 2), (1, 3)$
0%
$(-2, 2), (5, -7), (2, 9)$
0%
$(1, -3), (2, -3), (2, -7)$

Explanation

$2y-x=8$

$\implies x=2y-8$

Let,

$y=2$ , then

$x=2(2)-8=-4$ ,

$y-2x=1$

$\implies y=1+2x$

Let,

$x=1$ , then

$y=1+2(1)=3$

Similarly, when

$x=2,y=5$

and

$5y-x=14$

$\implies x=5y-14$

let,

$y=3$ then

$x=5(3)-14=1$

Similarly, when,

$y=2,x=-4$

Plotting the above points on the graph, we get intersection points

$A(-4,2), B(2,5),C(1,3)$

Solve graphically the pair of linear equations:

$4x - 3y + 4 = 0, 4x + 3y - 20 = 0$ . Find the area of the region bounded by these lines and

$x$ -axis

0%
$12$ sq. units
0%
$18$ sq. units
0%
$9$ sq. units
0%
$13$ sq. units

Explanation

$4x-3y+4=0$

$\implies x=\dfrac{-4+3y}{4}$

Let,

$y=4$ , then

$x=\dfrac{-4+3(4)}{4}=2$

Similarly, when

$y=0, x=-1$

$4x+3y-20=0$

$\implies x=\dfrac{20-3y}{4}$

Let,

$y=4$ , then

$x=\dfrac{20-3(4)}{4}=2$

Similarly, when

$y=0,x=5$

Plotting the above points on the graph, we get, the intersecting point i.e

$(2,4)$

Area of the region bounded by these lines and x-axis

$=\dfrac{1}{2}\times 6 \text{units} \times 4 \space \text{units}$

$=12 \space\text{ sq. units.}$

Based on equations reducible to linear equations
Solve for x and y:

$\dfrac {2}{x-1}+\dfrac {y-2}{4}=2; \dfrac {3}{2(x-1)}+\dfrac {2(y-2)}{5}=\dfrac {47}{20}$

0%
$x = 3, y = 6$
0%
$x = 1, y = 5$
0%
$x = 2, y = -7$
0%
$x = 5, y = -9$

Explanation

The given equations are
$\Rightarrow \dfrac{2}{x-1}+\dfrac{y-2}{4}=2......eq(1)$
$\Rightarrow \dfrac{3}{2\left (x-1 \right )}\dfrac{2\left (y-2 \right )}{4}=\dfrac{47}{20}......eq(2)$
$Put \dfrac{1}{x-1}=u \ and \ y-2=v \ in \ eq(1)\ and \ eq(2)$
$\Rightarrow 2u+\dfrac{v}{4}=2\Rightarrow 8u+v=8....eq(3)$
$\Rightarrow \dfrac{3u}{2}+\dfrac{2v}{5}=\dfrac{47}{20}\Rightarrow 30u+8v=47.....eq(4)$
$Multiply \ eq(3) \ by \ 8$
$\Rightarrow 64u+8v=64.....eq(5)$
$Subtract \ eq(4) \ and \ eq(5)$
$\Rightarrow \left (64u+8v=64 \right )-\left (30u+8v+47 \right )$
$\Rightarrow 34u=17\Rightarrow u=\dfrac{1}{2}$
$put \ u=\dfrac{1}{2} \ in \ eq(3)$
$\Rightarrow 8\times \dfrac{1}{2}+v=8\Rightarrow v=4$
$Hence, u=\dfrac{1}{x-1} =\dfrac{1}{2}\Rightarrow x-1=2\Rightarrow x=3$
$v=y-2=4\Rightarrow y=6$

Based on equations reducible to linear equations, solve for

$x$ and

$y$ :

$\dfrac {x-y}{xy}=9; \dfrac {x+y}{xy}=5$

0%
$x = -\dfrac12, y = \dfrac17$
0%
$x = -\dfrac15, y = \dfrac12$
0%
$x = -\dfrac15, y = \dfrac17$
0%
None of these

Explanation

The given equations are
$\Rightarrow \dfrac{x-y}{xy}=9\Rightarrow x-y=9xy....eq\left (1 \right )$
$\Rightarrow \dfrac{x+y}{xy}=5\Rightarrow x+y=5xy....eq\left (2 \right )$
Divide both eq by $xy$
$\Rightarrow \displaystyle \frac{1}{y}-\frac{1}{x}=9....eq\left ( 3 \right )$
$\Rightarrow \displaystyle \frac{1}{y}+\frac{1}{x}=5....eq\left ( 4 \right )$
Put $\dfrac{1}{x}=u$ and $\dfrac{1}{y}=v$ in eq $\left (3 \right )$ and $\left ( 4 \right )$
$\Rightarrow v-u=9 ....eq\left ( 5 \right )$
$\Rightarrow v+u=5 ....eq\left ( 6\right )$
Subtract eq $\left (5 \right )$ and eq $\left ( 6 \right )$
$\Rightarrow \left (v-u=9 \right )-\left ( v+u=5 \right )$
$\Rightarrow 2v=-14\Rightarrow v=7$
Put $v=7$ in eq $\left ( 5\right )$
$\Rightarrow 7+u=5\Rightarrow u=-2$
Hence, $\dfrac{1}{x}= u=-2\Rightarrow x=-\dfrac{1}{2}$
$\dfrac{1}{y}= v=7\Rightarrow y=\dfrac{1}{7}$

Based on equations reducible to linear equations
Solve for x and y

$\dfrac {1}{3x}-\dfrac {1}{7y}=\dfrac {2}{3}; \dfrac {1}{2x}-\dfrac {1}{3y}=\dfrac {1}{6}$

0%
$x = 1/8, y = 1/2$
0%
$x = 1/3, y = 1/6$
0%
$x = 1/7, y = 1/8$
0%
$x = 1/5, y = 1/7$

Explanation

The given equations are

$\Rightarrow \dfrac{1}{3x}-\dfrac{1}{7y}=\dfrac{2}{3}....eq\left ( 1 \right )$

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

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

$\dfrac{1}{y}=v$ in

$eq\left ( 1 \right )$ and

$eq \left (2 \right )$

$\Rightarrow \dfrac{1}{3}u-\dfrac{1}{7}v=\dfrac{2}{3}\Rightarrow 7u-3v=14....eq\left ( 3 \right )$

$\Rightarrow \dfrac{1}{2}u-\dfrac{1}{3}v=\dfrac{1}{6}\Rightarrow 3u-2v=1....eq\left (4 \right )$

$Substract\ eq\left ( 3 \right ) and\ eq\left ( 4 \right )$

$\Rightarrow 5u=25\Rightarrow u=5$

$put\ the\ value\ of\ u\ in\ eq\left ( 3 \right )$

$\Rightarrow 7\times 5-3v=14\Rightarrow v =7$

$Hence, \dfrac{1}{x}=u=5\Rightarrow x=\dfrac{1}{5}$

$\dfrac{1}{y}=v=7\Rightarrow y=\dfrac{1}{7}$

$4$ tables and

$3$ chairs together cost Rs.

$2250$ and

$3$ tables and

$4$ chairs cost Rs.

$1950$ . Find the cost of

$2$ chairs and

$1$ table.

0%
Rs. $150$
0%
Rs. $750$
0%
Rs. $350$
0%
Rs. $850$

Explanation

Suppose the cost of $1$ table $=x$ and cost of $1$ chair $=y$
Then according to the question
$\Rightarrow 4x+3y=2250$ ...........(1)
$\Rightarrow 3x+4y=1950$ .............(2)

Multiply (1) by $4$ and (2) by $3$ and Subtract both
$\Rightarrow (16x+12y=9000)- (9x+12y=5850)$

$\Rightarrow 7x=3150\Rightarrow x=450$

Put $x=450$ in $eq2$

$\Rightarrow 450\times4+3y=2250\Rightarrow 3y=2250-1800\Rightarrow 3y=450$

$\Rightarrow y=150$

Cost of $1$ table $=450$
Cost of $1$ chair $=150$
$\therefore$ Cost of $1$ table and $2$ chair $=450+150\times2=750$

A man has only

$20$ paise coins and

$25$ paise coins in his purse. If he has

$50$ coins in all totaling Rs.

$11.25$ , how many coins of each kind does he have?

0%
$25$ coins of each kind
0%
$16$ coins of each kind
0%
$44$ coins of each kind
0%
$32$ coins of each kind

Explanation

Let no of $20$ p coins $=x$
No of $25$ p coins $=y$
According to the question
$x+y=50$
$\Rightarrow x=50-y.....eq1$
And $20x+25y=1125\Rightarrow 4x+5y=225....eq2$
Put the value of $x$ from $eq1$
$\Rightarrow 4\left ( 50-y \right )+5y=225......eq3$
$\Rightarrow 200-4y+5y=225\Rightarrow y=25$
Put $y=25$ in $eq1$
$\Rightarrow x+25=50\Rightarrow x=25$

No of $20$ p coins $=25$

No of $25$ p coins $=25$

The sum of two numbers is

$8$ . If their sum is

$4$ times their difference. Find the numbers.

0%
$4, 4$
0%
$6, 2$
0%
$5, 3$
0%
None of these

Explanation

Let the first number be $x$ and second number $y.$

According to question
$x+y=8\quad\quad\quad\dots(i)$

$4(x-y)=8$

$\Rightarrow x-y=2\quad\quad\quad\dots(ii)$

Add equations $(i)$ and $(ii),$
$\begin{aligned}{}\left( {x + y} \right) + \left( {x - y} \right)& = 8 + 2\\2x &= 10\\x &= 5\end{aligned}$

Substitute $x=5$ in $(i),$
$\begin{aligned}{}\left( 5 \right) + y &= 8\\y&=8-5\\y &= 3\end{aligned}$

So, the numbers are $5$ and $3.$

The sum of the numerator and denominator of a fraction is

$4$ more than twice the numerator. If the numerator and denominator are increased by

$3$ , they are in the ratio

$2 : 3$ . Determine the fraction

0%
$\dfrac {1}{7}$
0%
$\dfrac {2}{5}$
0%
$\dfrac {3}{7}$
0%
$\dfrac {5}{9}$

CBSE Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 9 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 9 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

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