Class 12 Commerce Applied Mathematics - Linear Equations

Find the solution of given pair of linear equation by elimination method.
$3x+4y=-17 \rightarrow (1)$
$5x+2y=-19 \rightarrow (2)$

$3x+4y=-17$ (1)

$5x+2y=-19$ (2)

Multiplying (2) by 2

$10x+4y=-38$ (3)

Subtracting (1) from (3)

$7x=-21$

$x=-3$ (4)

Substituting (4) in (1)

$3(-3)+4y=-17$

$4y=-8$

$y=-2$

Hence the solution is

$x=-3$ and

$y=-2$ .

A man starts his job with a certain monthly salary and earns a fixed increment every year. If his salary was Rs $1500$ after $4$ year of service and Rs $1800$ after $10$ years of service, what was his starting salary and what is the annual increment?

Let the starting salary of the man be Rs.

$x$ and the fixed annual increment be Rs.

$y$ . Then,

Salary after

$4$ years of service

$=$ Rs.

$(x+4y)$
Salary after

$10$ years of service

$=$ Rs.

$(x+10y)$

$\therefore x+4y=1500$ (i)

$x+10y=1800$ (ii)

Substracting equation (i) from equation (ii), we get

$6y=300\Rightarrow y=50$

Putting

$y=50$ in equation (i), we get

$x=1300$ .

Hence the starting salary was Rs.

$1300$ and annual increment is Rs.

$50$ .

The value of $x$ and $y$ , if $\dfrac{1}{x-2y}=\dfrac{1}{3}$ and $\dfrac{1}{x+2y}=\dfrac{1}{7}$ is 5,1 respectively.Is it true ?If true enter 1 else 0.

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

$\dfrac{1}{x+2y}=\dfrac{1}{7}$ ---(2)

$\implies$

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

$x+2y=7$ ---(2)

_____________

$2x=10$

$\therefore x=5$

On substituting the value of

$x$ in equ

$1$ .

$5-2y=3$

$5-3=2y$

$\therefore y=1$

so given statement is true.

If $2x+y=9$ and $3x-y=6$ then find the value of $x$ .

Given,

$2x+y=9$ --------------------------

$(1)$

$3x-y=6$ --------------------------

$(2)$

Adding

$e{q^n}\left( 1 \right)$ and

$e{q^n}\left( 2 \right)$

$2x+y=9$

$+$

$(3x-y=6)$

$=9+6$

$x=\dfrac { { 15 } }{ 5 } =3$

We get,

$x=3$

Solve: $3x+4y=10,2x-2y=2$ by the method of elimination.

$3x+4y=10$ …………..

$(1)$

$2x-2y=2$ (or)

$x-y=1$ …………

$(2)$

$(1)$

$\Rightarrow (3x+4y=10)1$

$(2)$

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

_____________________

$3x+4y=10$

$3x-3y=3$

_____________________

$7y=7$

$\therefore y=1$

Put

$y=1$ in

$(1)$ we get

$3x+4\times 1=10$

$3x=10-4=6$

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

Hence

$x=2, y=1$ .

1257517_1328052_ans_af5eae8fa2eb4f97a3206f6ec031bb26.PNG

Solve the following
$3 a + 5 b = 26 ; a + 5 b = 27$

$3a+5b=26$

$-a-5b=-27$

__________

$2a=-1$ (on subtracting)

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

$5b=27-a$

$5b=27+\dfrac{1}{2}=\dfrac{55}{2}$

$b=\dfrac{11}{2}$

$\therefore a=\dfrac{-1}{2}, \ b=\dfrac{11}{2}$

Solve the following simultaneous equations
$3 a + 5 b = 20 ; a + 5 b = 22$

$3a+5b =20$

$a+5b=22$

subtracting both the equations, we get

$3a-a=20-22$

$2a=-2$

$a=-1$

$a + 5b = 22$

$5b = 22 - a = 22 + 1 = 23$

$b= \dfrac {23}{5}$

Hence

$a = - 1$ and

$b= \dfrac{23}{5}$

Solve the following simultaneous equations
2x+y=5; 3x-y=5

We have

$2x+y=5$

$...(1)$

$3x-y=5$

$...(2)$

Adding equations

$(1)$ and

$(2),$

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

$\Rightarrow 5x=10$

$\Rightarrow x=2$

Substitute the value of

$x$ in equation

$(1)$

$\Rightarrow y=5-2\times 2$

$\Rightarrow y=5-4$

$\Rightarrow y=1$

$\therefore$ The value of

$x=2$ and

$y=1$

Solve the following
$2 x - 3 y = 9 ; 2 x + y = 13$

$2x - 3y = 9$

$2x + y = 13$

subtracting , we get

$y + 3y = 10 - 9$

$4y=4$

$\Rightarrow y=1$

$2x - 3y = 9$

$2x = 9 + 3 = 12$

$x = 6$

$\therefore x=6, \ y=1$

$\cfrac{x}{3}+\cfrac{y}{4}=4;\ x+y=14$

Given equation,

$\dfrac { x }{ 3 } +\dfrac { y }{ 4 } =4,\quad x+y=14$

$\Rightarrow 4x+3y=48\quad \longrightarrow \left( 1 \right) ,\quad 4x+4y=56\longrightarrow \left( 2 \right)$

Subtracting equation

$(1)$ and

$(2)$

$4x+3y-48=0\\ 4x+4y-56=0\\ --------\\ 0-y+8=0$

$y=8$

For

$y=8, x=6$

$(x, y)=(6, 8)$

$5y-3x=14; \, \, 3y-2x=1$

Given equations are,

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

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

Multiply (1) by

$2$ and (2) by

$3$ we get,

$10y-6x=28$ .........(3)

$9y-6x=3$ ..........(4)

Subtracting (4) from (3) we get,

$(10y-6x)-(9y-6x)=28-3$

$\Rightarrow y=25$

Substituting

$y=25$ in (1) we get,

$5\times 25-3x=14$

$\Rightarrow 125-3x=14$

$\Rightarrow -3x=14-125$

$\Rightarrow -3x=-111$

$\Rightarrow x=37$

Hence, the value of

$x$ is

$37$

Solve the following pair of linear equations by the elimination method and find the value of $p+q$ rounded off to the nearest integer.
$4p=3q+5; \, \, p-q = \dfrac{7}{6}$

$4p=3q+5\quad\quad\quad\dots(i)$

$p-q = \dfrac{7}{6}\quad\quad\quad\dots(ii)$

Rewriting

$(i)$ we get,

$4p-3q=5\quad\quad\quad\dots(iii)$

Multiplying

$(ii)$ by

$4$ we get,

$4p-4q = \dfrac{14}{3}\quad\quad\quad\dots(iv)$

Subtracting

$(iv)$ from

$(iii)$ we get,

$\begin{array}{l}\underline {\begin{array}{ccccccccccccccc}{4p}& - &{3q}& = &5\\{4p}& - &{4q}& = &{\dfrac{{14}}{3}}\\ - & + &{}&{}& - \end{array}} \\\;\;\;\;\;\quad\begin{array}{ccccccccccccccc}{}&{}&q& = &{\dfrac{1}{3}}\end{array}\end{array}$

Re substituting

$q=\dfrac{1}{3}$ in

$(i)$ we get,

$\begin{aligned}{}4p &= 3\left( {\frac{1}{3}} \right) + 5\\4p &= 1 + 5\\4p &= 6\\p &= \frac{3}{2}\end{aligned}$

So,

$\begin{aligned}{}p + q &= \frac{3}{2} + \frac{1}{3}\\ &= \frac{{9 + 2}}{6}\\ &= \frac{{11}}{6}\\ &\approx 2\end{aligned}$

$2x-y-3=0; \, \, 4x-y-5=0$

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

$4x-y=5$ ......(2)
Subtracting (1) and (2) we get,

$x=1$
Re substituting

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

$y=-1$

$\displaystyle \frac{3}{2}x + \frac{2}{3}y = 13 \frac{1}{3}; \, \, \frac{2}{3}x + \frac{3}{2}y = 8 \frac{1}{3}$

Given,

$\displaystyle \frac{3}{2}x + \frac{2}{3}y = 13 \frac{1}{3}$ .....(1)

$\displaystyle \frac{2}{3}x + \frac{3}{2}y = 8 \frac{1}{3}$ .......(2)

Rewriting above equations we get,

$9x+4y=80$ .......(3)

$4x+9y=50$ ........(4)

Multiplying (3) by 4 and (4) by 9 we get,

$36x+16y=320$ .......(5)

$36x+81y=450$ ........(6)

Subtracting (5) and (6) we get,

$y=2$
Re substituting

$y=2$ in (5) we get,

$x=8$

$\therefore$

$x+y=8+2=10$

$3x-2y = 4; \, \, 6x+7y=19$

$3x-2y = 4$ ..........(1)

$6x+7y=19$ .........(2)
Multiply (1) by 2 we get,

$6x-4y = 8$ ..........(3)
Subtracting (2) and (3) we get,

$y=1$
Re substituting

$y=1$ in (1) we get,

$x=2$

Solve for $x$ & $y$

$ax+by=a-b$
$bx-ay=a+b$

The given set of equations is,

$ax + by = a - b$ ....(1)

$bx - ay = a + b$ ....(2)

Multiplying (1) by $b$ and (2) by $a$ and subtracting (2) from (1), we get:

$\left( {{a^2} + {b^2}} \right)y = - \left( {{a^2} + {b^2}} \right)$

$y = - 1$

Substituting $y = - 1$ in (1), we get:

$ax + b\left( { - 1} \right) = a - b$

$ax - b = a - b$

$ax = a$

$x = 1$

Therefore, the solution of the given system of equations is $x = 1,\ y = - 1$ .

Find the fraction which becomes to $\dfrac23$ when the numerator is increasing by $2$ and equal to $\dfrac47$ when the denominator is increased by $4$

Consider that the fraction is $\dfrac{x}{y}$ .

According to the question,

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

$3(x+2)=2y$

$3x - 2y = - 6$ .............. (1)

And,

$\dfrac{x}{{y + 4}} = \dfrac{4}{7}$

$7x=4(y+4)$

$7x - 4y = 16$ ............. (2)

From equation (1) and (2),
$2 \times (1) - (2) \implies$

$6x - 4y = - 12$

$7x - 4y = 16$

$..............$

$-x =-28$

$\implies x = 28$

in $(1)$ , we get
$3(28)-2y =-6$

$2y =90$

$y = 45$

Therefore, the fraction becomes $\dfrac{{28}}{{45}}$ .

Solve the following pair of linear equations in two variables:
$3=2x+y$ and $9 = 4x - y$

Given:-

$3 = 2x + y ..... \left( 1 \right)$

$9 = 4x - y ..... \left( 2 \right)$

Adding equation

$\left( 1 \right) \& \left( 2 \right)$ , we have

$3 + 9 = \left( 2x + y \right) + \left( 4x - y \right)$

$12 = 6x$

$\Rightarrow x = \cfrac{12}{6} = 2$

Substituting the value of

$x$ in equation

$\left( 1 \right)$ , we have

$3 = 2 \times 2 + y$

$\Rightarrow y = 3 - 4 = -1$

Hence the value of

$x$ and

$y$ are

$2$ and

$-1$ respectively.

Solve for $x$ and $y:x-4y-14=0$ , $5x-y-13=0$

Given: Equation of lines

$x-4y-14=0$ and

$5x-y-13=0$

$x-4y=14$ .....

$(1)$

$5x-y=13$ .....

$(2)$

Multiplying eq 1 by 5 then subtracting eq 2 from the resultant multiplication i.e,

$5\times (1)-1\times (2)$ we get;

$5x-20y-5x+y=70-13$

$\Rightarrow -19y=57$

$\Rightarrow y=\dfrac{57}{-19}=-3$

Putting

$y=-3$ in

$(1)$ we get;

$x-4\times -3=14$

$\Rightarrow x+12=14$

$\Rightarrow x=14-12=2$

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

Solve using elimination method :
$7x + 2y = 19$
$x - y = 4$

Given equations are

$7x + 2y = 19......(1)$

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

Multiplying equation

$(2)$ by

$2$

$2x - 2y = 8......(3)$

Now, adding equation

$1$ and equation

$2$ ,

$(7x+2y)+(2x-2y)=19+8$

$\Rightarrow 7x+2y+2x-2y=27$

$\Rightarrow 9x=27$

$\Rightarrow x=3$

Putting

$x=3$ in

$(2)$ we get,

$3-y=4$

$\Rightarrow y=3-4$

$\Rightarrow y=-1$

Hence, the solution is

$x=3$ and

$y=-1$ .

Solve the following system of equation for x and y.
$\displaystyle \frac{5}{x-1}+\frac{1}{y-2}=2,\frac{6}{x-1}-\frac{3}{y-2}=1$

Let

$\dfrac{1}{x-1}=a$ and

$\dfrac{1}{y-2}=b$

$5a+b=2 ..... (1) .... \times 3$

$6a-3b=1 ..... (2) ...... \times 1$

$15a+3b=6$

$6a-3b$

$\_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_$

$21a=7$

$\therefore \boxed{a=\dfrac{1}{3}}$

Putting

$a=\dfrac{1}{3}$ in

$(1)$

$\dfrac{5}{3}+b=2$

or,

$b=2-\dfrac{5}{3}=\dfrac{1}{3}$

Substituting value of

$a$ and

$b$

$\dfrac{1}{x-1}=\dfrac{1}{3}\Rightarrow x-1=3 ..... (3)$

$\dfrac{1}{y-2}=\dfrac{1}{3}\Rightarrow y-2=3 ...... (4)$

From

$(3), x=4$

From

$(4), y=5$

Solve : $\dfrac{x}{a}+\dfrac{y}{b}=2$
$ax-by=a^{2}-b^{2}$

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

or,

$ba+ay=2ab .... (1) .....\times a$

$ax-by=a^{2}-b^{2} ..... (2).... \times b$

$abx+a^{2}y=2a^{2}b$

$aba-b^{2}y=a^{2}b-b^{3}$

$\_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_$

$(a^{2}+b^{2})y=(a^{2}+b^{2})b$

$\therefore y=b$

Putting

$y=b$ in

$(1)$

$bx+ab=2ab$

or,

$bx=ab$

$\therefore x=a$

Solve the following systems of linear equations

$i)2x - y = 17$

$ii)3x + 5y = 6$

Given equations are

$2x-y=17$ ............(i)

$3x+5y=6$ ...........(ii)

$10x-5y=85$ ............multiplying eq. (i) with

$5$ .........(iii)

(ii)+(iii)

$13x=91$

$x=7$

Putting value of

$x$ in (ii)

$14-y=17$

$\Rightarrow y=-3$

Solve for $x$ and $y$ :
$2x-y+3=0$ ,
$3x-7y+10=0$ .

$2x-y+3=0 ...(1)$

$3x-7y+10=0 ....(2)$

Let us use elimination method to solve the given system of equations.
Find the value of

$y$ from equation

$(1)$ and substitute the value in

$(2)$ .

From

$(1)$

$-y=-3-2x\Rightarrow y=2x+3$
And

$3x-7(2x+3)=-10\Rightarrow 3x-14x-21=-10$

$-11x=-10+21=11\Rightarrow x=\dfrac {11}{-11}=-1$
and

$y=2x+3$

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

$y=1$

Answer :

$x=-1$ and

$y=1$

Solve for $x$ and $y$ :
$3x-5y-19=0$ ,
$-7x+3y+1=0$ .

$3x-5y-19=0 ...(1)$

$-7x+3y+1=0 ....(2)$

Let us use elimination method to solve the given system of equations.

Multiply

$(1)$ by

$3$ and

$(2)$ by

$5$ . And add both the equations.

$\dfrac { 9x-15y=57\\ -35x+15y=-5 }{ -26x=52 }$

$x=\dfrac {52}{-26}= -2$

Substitute the value of

$x$ in equation

$(1)$ , we have

$3\times (-2)-5y=19\Rightarrow -6-5y=19$

$y=-5$

Hence

$x=-2$ and

$y=-5$

$\Rightarrow x=-2, y=-5$

Solve for $x$ and $y$ :
$x/3+y/4=11$ ,
$5x/6-y/3=-7$

$x/3+y/4=11$ ,

$5x/6-y/3=-7$

Simplify the both equations:

$\dfrac {x}{3}+\dfrac {y}{4}=11\Rightarrow {4x+3y=132}$ and

$\dfrac {5x}{6}-\dfrac {y}{3}=-7 \Rightarrow {5x-2y=-42}$

We have,

$4x+3y=132 ...(1)$

$5x-2y=-42 ...(2)$

Let us use elimination method to solve the given system of equation.
Multiply

$(1)$ by

$2$ and

$(2)$ by

$3$ . And add both the equations.

$\dfrac { 8x+6y=264\\ 15x-6y=-126 }{ 23x\quad =138 }$

$x=\dfrac {138}{23}=6$

From

$(1): 4(6)+3y=132$

$y=108/3 =36$

Answer

$x=6$ and

$y=36$

Solve for $x$ and $y$ :
$2x+3y=0$ ,
$3x+4y=5$ .

$2x+3y=0 ....(1)$

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

Let us the use elimination method to solve the given system of equations.
Multiply

$(1)$ by

$4$ and

$(2)$ by

$3$ . And subtract both the equations.

$\dfrac { 8x+12y=0\\ 9x+12y=15\\ -\quad -\quad - }{ -x\quad =-15\Rightarrow x=15 }$

From

$(1), y=-10$

Hence

$x=15$ and

$y=-10$

The linear equation that converts Fahrenheit $(F)$ to Celsius $(C)$ is given by the relation:
$C=\dfrac{5F - 160}{9}$
(i) If the temperature is $86 F$ , what is the temperature in Celsius?
(ii) If the temperature in $35$ C, what is the temperature in Fahrenheit?
(iii) If the temperature is $0$ C, what is the temperature in Fahrenheit and if the temperature is 0 F, what is the temperature in Celsius?
(iv) What is the numerical value of the temperature which is same in both the scales?

$C=\dfrac{5F - 160}{9}$
(i) Putting

$F = 86^{\circ }$ , we get

$C = \dfrac{5(86)-160}{9}=\dfrac{430-160}{9}=\dfrac{270}{9}=30^{\circ }$

Hence, the temperature in Celsius in

$30^{\circ } C.$

(ii) Putting C =

$35^{ \circ }$ , we get

$35^{\circ } = \dfrac{5(F)-160}{9}$

$\Rightarrow 315^{\circ } = 5F - 160$

$\Rightarrow 5F = 315+160=475$

$\therefore F=\dfrac{475}{5} = 95^{\circ }$

Hence, the temperature in Fahrenheit is

$95^{\circ }$ F.

(iii) Putting C =

$0^{\circ }$ , we get

$0 = \dfrac{5F-160}{9} \Rightarrow 0 = 5F - 160$

$\Rightarrow 5F = 160$

$\therefore F = \dfrac{160}{5} = 32^{\circ }$
Now, putting F =

$0^{\circ }$ , we get

$C = \dfrac{5F-160}{9} \Rightarrow C = \dfrac{5(0)-160}{9} =\left ( -\dfrac{160}{9} \right )^{\circ }$

If the temperature is

$0^{\circ }$ C, the temperature in Fahrenheit is

$32^{\circ }$ and if the temperature is

$0^{\circ }$ F, then the temperature in Celsius is

$\left ( -\dfrac{160}{9} \right )^{\circ }$ C.

(iv) Putting

$C = F,$ in the given relation, we get

$F = \dfrac{5F-160}{9} \Rightarrow 9F = 5F - 160$

$\Rightarrow 4F = - 160$

$\therefore F = \dfrac{-160}{4} = -40^{\circ }$
Hence, the numerical value of the temperature which is same in both the sacales is

$-40.$

Solve the following pair of linear equations by the elimination method :
3x - 5y - 4 = 0 and 9x = 2y + 7

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

$9x - 2y = 7 ........(ii)$
Multiplying eqn (i) by 3

$9x - 15 y = 12........(iii)$

subtract eqn (iii) from (ii

$13y=-5$

$y=-\dfrac{5}{13}$

Put in eqn (i)

$3x - 5\times\dfrac{-5}{13} = 4$

Multiplying by

$13$ on both side

$39x+25=52$

$39x=27$

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

From the pair of linear equations in the following problems , and find their solutions ( if they exist ) by the elimination method :
The sum of the digits of a two - digit number is 9 . Also , nine times this number is twice the number obtained by reversing the order of the digits . Find the number :

Let the two digit number be

$10x+y$

Given,

$x+y=9 .....(1)$

And,revers number is

$10y+x$

$9(10x+y)=2(10y+x)$

$90x+9y=20y+2x$

$88x−11y=0$ .....(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

$(1),$

$1+y=9$

$y=8$

Then the number is

$10x+y=10(1)+8=10+8=18$

Hence, the number is 18.

From the pair of linear equations in the following problems , and find their solutions ( if they exist ) by the elimination method :
A leading library has a fixed charge for the first three days and 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 .

Let the fixed charge be

$Rs. x$

and the extra charge be

$Rs. y$

$x+4y=27⟶(i)$

$x+2y=21⟶(ii)$

On substracting (ii) from (i),

$2y=6$

$y=3$

putting y in (i),

$x+4(3)=27$

$x=15$

$x=Rs.15$ and

$y=Rs. 3$

Fixed charge

$=Rs. 15$

Charge for each extra day

$=Rs. 3$ .

From the pair of linear equations in the following problems , and find their solutions ( if they exist ) by the elimination method :
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 .

Let the Rs . 50 notes be 'x'
Rs 100 notes be 'y'

$x+ y = 25.....(i)$

$50x+100y=2000$

$x + 2y = 40.....(ii)$

Subtracting eqn (i) from (ii)

$y = 15$

Put in eqn(i)

$x + y = 15$

$x + 15 = 25$

$x = 25 - 15$

$x = 10$

Let the

$Rs . 50$ notes be

$'10'$
Rs

$100$ notes be

$'15'$

Solve the following pair of linear equations by the elimination method the substitution method :
3x + 4 y = 10 and 2x - 2y = 2

$3x + 4 y = 10$ and

$2x - 2y = 2$

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

$2x - 2y = 2 ............(ii)$
Multiplying eqn (ii) by 2,

$4x - 4y = 4 ...........(iii)$
Adding eqn (i) to eqn (iii),

$7x = 14$

$x=2$

Put in eqn (i)

$3x + 4 y = 10$

$3(2) + 4y = 10$

$4y = 10 - 6$

$4y = 4$

$y=1$

$x=2,y=1$

Solve the following pair of linear equations by the elimination method the substitution method :
$\dfrac{x}{2} + \dfrac{2y}{3} = -1 and x - \dfrac{y}{3} = 3$

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

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

$\dfrac{3x + 4y}{6} = -1\,\,\,\,\,\, \dfrac{ 3x - y}{3} = 3$

$3x + 4y = - 6 ...(i)$

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

Subtract eqn (ii)from (i)

$5y=-15$

$y = \dfrac{-15}{5} = - 3$

$y = -3$

$3x + 4y = -6$

$3x + 4 (-3) = - 6$

$3x - 12 = - 6$

$3x = - 6 + 12$

$3x = 6$

$\therefore x = 2 , y = 3$

Solve the following pair of linear equations by suitable method
x + y = 5 and 2x - 3y = 4

$x + y = 5$ and

$2x - 3y = 4$

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

$2x - 3y = 4 .............(ii)$
Multiplying eqn (i) by 3

$3x + 3y = 15 ..........(iii)$
By adding

$(ii)$ to

$(iii)$ , '

$y$ ' is eliminated .

$5x=19$

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

$x + y = 5$

$\dfrac{19}{5} + y = 5$

$y = \dfrac{5}{1} - \dfrac{19}{5}$

$= \dfrac{25 - 19}{5}$

$y=\dfrac{6}{5}$

Solve the following pair of linear equations
$( a - b) x + ( a + b) y = a^{2} - 2ab - b^{2} \rightarrow (1)$
$( a + b) (x + y) = a^{2} + b^{2}$

$( a - b) x + ( a + b) y = a^{2} - 2ab - b^{2} \rightarrow (1)$

$( a + b) (x + y) = a^{2} + b^{2}$

$(a + b) x + ( a + b)y= a^{2} + b^{2} \rightarrow (2)$

using image

$-2bx=-2ab-2b^2$

$x=a+b$

putting in (2)

$(a+b)^2 +(a+b)y=a^2+b^2$

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

Hence Required answer

1959858_1867786_ans_6c22e5cc56c046cb951cb7d65fce73f0.png

Solve the following pair of the equations by reducing them to a pair of linear equations:
$6x + 3y = 6xy$
$2x + 4y = 5xy$

$6x + 3y = 6xy$ ...............(A)

$2x + 4y = 5xy$ .............(B)

Dividing by

$xy$ in LHS and RHS of both the equations we get,

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

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

Let

$u=\dfrac{1}{x},v=\dfrac{1}{y}$ we get,

$6v + 3u = 6$

$2v + 4u = 5$

$6v + 3u = 6\rightarrow (1)$

$2v + 4u = 5\rightarrow (2)$

Multiplying eq. (2) by

$3$ and Subtracting from (1) we get,

$6v-6v + 3u-12u=6-15$

$-9u=-9$

$u=1$

$\therefore v=\dfrac{6-3}{6}=\dfrac{1}{2}$ {Using eq. (1)}

$\therefore x=1,y=2$

From the pair of linear equation in the following problems and find their solutions (if they exist ) by any algebraic method :
The area of a rectangular gets reduced by 9 square units , its length is reduced by 5 nits and breadth in increased by 3 units . If we increase the length by 3 units and the breadth by 2 units , the area increases by 67 square units . Find the dimensions of the rectangle .

$Length\space of\space rectangular\space is\space "x"\space unit$

$Breadth\space of\space rectangular\space is\space "y" unit$

$Original \space area=xy$

$According \space to \space the \space first \space condition$

$( x- 5 ) ( y + 3 ) = xy - 9$

$xy + 3x - 5y - 15 = xy - 9$

$3x - 5y = -9 + 15$

$3x - 5y = 6.........(1)$

$According \space second \space condition$

$(x+3)(y+2)= xy+67$

$xy+2x+3y+6=xy+67$

$2x+3y=61........(2)$

$Multiply \space eq(1) \space by\space 2 \space and \space eq(2) \space by \space 3\space and \space subtracting \space we \space get$

$Solving \space we \space get$

$x=17 \space and \space y=9$

Solve the following pair of linear equations
$px + qy = p - q$
$qx - py = p + q$

$px + qy = p - q.......(1)$

$qx - p y = p + q..........(2)$

$Multiplying \space eqn(1 )\space p\space and\space eqn(2)\space by\space q$

$p^{2}x + pqy = p^{2} - pq ........(3)$

$q^{2}x - pqy = pq + q^{2}........(4)$

$Adding\space (3)\space and\space (4)$

$p ^{2}x +q^{2}x = p^{2}+q^2$

$x(p^2+q^2)=p^2+q^2$

$\implies x=1$

$Multiplying \space eqn(1 )\space q\space and\space eqn(2)\space by\space p$

$pqx+q^2y=pq-q^2$

$pqx-p^2y=pq+q^2$

$Subtracting \space we \space get$

$(q^2+p^2)y=-(p^2+q^2)$

$\implies y=-1$

$x=1\space y=-1$

Solve the simultaneous equation and find the value of y
$\displaystyle \frac{x+y-8}{2} = \frac{x+2y-14}{3} = \frac{3x+y-12}{11}$

Consider two equations,

$\displaystyle \frac{x+y-8}{2} = \frac{3x+y-12}{11}$ ............(1)

$\displaystyle \frac{x+y-8}{2}= \frac{x+2y-14}{3}$ .............(2)

After solving (1) we get,

$5x+9y=64$ ..............(3)
After solving (2) we get,

$x-y=-4$
or

$9x-9y=-36$ ...............(4)
Adding (3) and (4) we get,

$14x=28$

$\therefore x=2$
Re substituting

$x=2$ in (3) we get,

$y=6$

Linear Equations - Class 12 Commerce Applied Mathematics - Extra Questions

Find the solution of given pair of linear equation by elimination method.
$3x+4y=-17 \rightarrow (1)$
$5x+2y=-19 \rightarrow (2)$

A man starts his job with a certain monthly salary and earns a fixed increment every year. If his salary was Rs $1500$ after $4$ year of service and Rs $1800$ after $10$ years of service, what was his starting salary and what is the annual increment?

The value of $x$ and $y$ , if $\dfrac{1}{x-2y}=\dfrac{1}{3}$ and $\dfrac{1}{x+2y}=\dfrac{1}{7}$ is 5,1 respectively.Is it true ?If true enter 1 else 0.

If $2x+y=9$ and $3x-y=6$ then find the value of $x$ .

Solve: $3x+4y=10,2x-2y=2$ by the method of elimination.

Solve the following
$3 a + 5 b = 26 ; a + 5 b = 27$

Solve the following simultaneous equations
$3 a + 5 b = 20 ; a + 5 b = 22$

Solve the following simultaneous equations
2x+y=5; 3x-y=5

Solve the following
$2 x - 3 y = 9 ; 2 x + y = 13$

$\cfrac{x}{3}+\cfrac{y}{4}=4;\ x+y=14$

$5y-3x=14; \, \, 3y-2x=1$

Solve the following pair of linear equations by the elimination method and find the value of $p+q$ rounded off to the nearest integer.
$4p=3q+5; \, \, p-q = \dfrac{7}{6}$

$2x-y-3=0; \, \, 4x-y-5=0$

$\displaystyle \frac{3}{2}x + \frac{2}{3}y = 13 \frac{1}{3}; \, \, \frac{2}{3}x + \frac{3}{2}y = 8 \frac{1}{3}$

$3x-2y = 4; \, \, 6x+7y=19$

Solve for $x$ & $y$

$ax+by=a-b$
$bx-ay=a+b$

Find the fraction which becomes to $\dfrac23$ when the numerator is increasing by $2$ and equal to $\dfrac47$ when the denominator is increased by $4$

Solve the following pair of linear equations in two variables:
$3=2x+y$ and $9 = 4x - y$

Solve for $x$ and $y:x-4y-14=0$ , $5x-y-13=0$

Solve using elimination method :
$7x + 2y = 19$
$x - y = 4$

Solve the following system of equation for x and y.
$\displaystyle \frac{5}{x-1}+\frac{1}{y-2}=2,\frac{6}{x-1}-\frac{3}{y-2}=1$

Solve : $\dfrac{x}{a}+\dfrac{y}{b}=2$
$ax-by=a^{2}-b^{2}$

Solve the following systems of linear equations

$i)2x - y = 17$

$ii)3x + 5y = 6$

Solve for $x$ and $y$ :
$2x-y+3=0$ ,
$3x-7y+10=0$ .

Solve for $x$ and $y$ :
$3x-5y-19=0$ ,
$-7x+3y+1=0$ .

Solve for $x$ and $y$ :
$x/3+y/4=11$ ,
$5x/6-y/3=-7$

Solve for $x$ and $y$ :
$2x+3y=0$ ,
$3x+4y=5$ .

Solve the following pair of linear equations by the elimination method :
3x - 5y - 4 = 0 and 9x = 2y + 7

Solve the following pair of linear equations by the elimination method the substitution method :
3x + 4 y = 10 and 2x - 2y = 2

Solve the following pair of linear equations by the elimination method the substitution method :
$\dfrac{x}{2} + \dfrac{2y}{3} = -1 and x - \dfrac{y}{3} = 3$

Solve the following pair of linear equations by suitable method
x + y = 5 and 2x - 3y = 4

Solve the following pair of linear equations
$( a - b) x + ( a + b) y = a^{2} - 2ab - b^{2} \rightarrow (1)$
$( a + b) (x + y) = a^{2} + b^{2}$

Solve the following pair of the equations by reducing them to a pair of linear equations:
$6x + 3y = 6xy$
$2x + 4y = 5xy$

Solve the following pair of linear equations
$px + qy = p - q$
$qx - py = p + q$

Solve the simultaneous equation and find the value of y
$\displaystyle \frac{x+y-8}{2} = \frac{x+2y-14}{3} = \frac{3x+y-12}{11}$

Class 12 Commerce Applied Mathematics Extra Questions

Linear Equations - Class 12 Commerce Applied Mathematics - Extra Questions

Find the solution of given pair of linear equation by elimination method.3x+4y=−17→(1)3x+4y=-17 \rightarrow (1)5x+2y=−19→(2)5x+2y=-19 \rightarrow (2)

A man starts his job with a certain monthly salary and earns a fixed increment every year. If his salary was Rs 15001500 after 44 year of service and Rs 18001800 after 1010 years of service, what was his starting salary and what is the annual increment?

The value of xx and yy, if 1x−2y=13\dfrac{1}{x-2y}=\dfrac{1}{3} and 1x+2y=17\dfrac{1}{x+2y}=\dfrac{1}{7} is 5,1 respectively.Is it true ?If true enter 1 else 0.

If 2x+y=92x+y=9 and 3x−y=63x-y=6 then find the value of xx.

Solve: 3x+4y=10,2x−2y=23x+4y=10,2x-2y=2 by the method of elimination.

Solve the following 3a+5b=26;a+5b=273 a + 5 b = 26 ; a + 5 b = 27

Solve the following simultaneous equations3a+5b=20;a+5b=223 a + 5 b = 20 ; a + 5 b = 22

Solve the following simultaneous equations2x+y=5; 3x-y=5

Solve the following 2x−3y=9;2x+y=132 x - 3 y = 9 ; 2 x + y = 13

Solve the following pair of equations:x3+y4=4; x+y=14\cfrac{x}{3}+\cfrac{y}{4}=4;\ x+y=14

Solve the following simultaneous equations by elimination method and find the value of xx 5y−3x=14;3y−2x=15y-3x=14; \, \, 3y-2x=1

Solve the following pair of linear equations by the elimination method and find the value of p+qp+q rounded off to the nearest integer.4p=3q+5;p−q=764p=3q+5; \, \, p-q = \dfrac{7}{6}

Solve the following simultaneous equations by elimination method and find the value of x+yx+y 2x−y−3=0;4x−y−5=02x-y-3=0; \, \, 4x-y-5=0

Solve the following simultaneous equations. and find x+yx+y for 32x+23y=1313;23x+32y=813\displaystyle \frac{3}{2}x + \frac{2}{3}y = 13 \frac{1}{3}; \, \, \frac{2}{3}x + \frac{3}{2}y = 8 \frac{1}{3}

Solve the following simultaneous equations by elimination method and find the value of xx 3x−2y=4;6x+7y=193x-2y = 4; \, \, 6x+7y=19

Solve for xx & yyax+by=a−bax+by=a-bbx−ay=a+bbx-ay=a+b

Find the fraction which becomes to 23\dfrac23 when the numerator is increasing by 22 and equal to 47\dfrac47 when the denominator is increased by 44

Solve the following pair of linear equations in two variables:3=2x+y3=2x+y and 9=4x−y9 = 4x - y

Solve for xx and y:x−4y−14=0y:x-4y-14=0, 5x−y−13=05x-y-13=0

Solve using elimination method :7x+2y=197x + 2y = 19x−y=4x - y = 4

Solve the following system of equation for x and y.5x−1+1y−2=2,6x−1−3y−2=1\displaystyle \frac{5}{x-1}+\frac{1}{y-2}=2,\frac{6}{x-1}-\frac{3}{y-2}=1

Solve : xa+yb=2\dfrac{x}{a}+\dfrac{y}{b}=2ax−by=a2−b2ax-by=a^{2}-b^{2}

Solve the following systems of linear equationsi)2x−y=17i)2x - y = 17ii)3x+5y=6ii)3x + 5y = 6

Solve for xx and yy:2x−y+3=02x-y+3=0,3x−7y+10=03x-7y+10=0.

Solve for xx and yy:3x−5y−19=03x-5y-19=0,−7x+3y+1=0-7x+3y+1=0.

Solve for xx and yy:x/3+y/4=11x/3+y/4=11,5x/6−y/3=−75x/6-y/3=-7

Solve for xx and yy:2x+3y=02x+3y=0,3x+4y=53x+4y=5.

Solve the following pair of linear equations by the elimination method : 3x - 5y - 4 = 0 and 9x = 2y + 7

Solve the following pair of linear equations by the elimination method the substitution method : 3x + 4 y = 10 and 2x - 2y = 2

Solve the following pair of linear equations by the elimination method the substitution method :x2+2y3=−1andx−y3=3 \dfrac{x}{2} + \dfrac{2y}{3} = -1 and x - \dfrac{y}{3} = 3

Solve the following pair of linear equations by suitable method x + y = 5 and 2x - 3y = 4

Solve the following pair of linear equations (a−b)x+(a+b)y=a2−2ab−b2→(1) ( a - b) x + ( a + b) y = a^{2} - 2ab - b^{2} \rightarrow (1) (a+b)(x+y)=a2+b2 ( a + b) (x + y) = a^{2} + b^{2}

Solve the following pair of the equations by reducing them to a pair of linear equations:6x+3y=6xy6x + 3y = 6xy 2x+4y=5xy2x + 4y = 5xy

Solve the following pair of linear equations px+qy=p−q px + qy = p - q qx−py=p+q qx - py = p + q

Solve the simultaneous equation and find the value of y x+y−82=x+2y−143=3x+y−1211\displaystyle \frac{x+y-8}{2} = \frac{x+2y-14}{3} = \frac{3x+y-12}{11}

Class 12 Commerce Applied Mathematics Extra Questions

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Find the solution of given pair of linear equation by elimination method.
$3x+4y=-17 \rightarrow (1)$
$5x+2y=-19 \rightarrow (2)$

A man starts his job with a certain monthly salary and earns a fixed increment every year. If his salary was Rs $1500$ after $4$ year of service and Rs $1800$ after $10$ years of service, what was his starting salary and what is the annual increment?

The value of $x$ and $y$ , if $\dfrac{1}{x-2y}=\dfrac{1}{3}$ and $\dfrac{1}{x+2y}=\dfrac{1}{7}$ is 5,1 respectively.Is it true ?If true enter 1 else 0.

If $2x+y=9$ and $3x-y=6$ then find the value of $x$ .

Solve: $3x+4y=10,2x-2y=2$ by the method of elimination.

Solve the following
$3 a + 5 b = 26 ; a + 5 b = 27$

Solve the following simultaneous equations
$3 a + 5 b = 20 ; a + 5 b = 22$

Solve the following simultaneous equations
2x+y=5; 3x-y=5

Solve the following
$2 x - 3 y = 9 ; 2 x + y = 13$

$\cfrac{x}{3}+\cfrac{y}{4}=4;\ x+y=14$

$5y-3x=14; \, \, 3y-2x=1$

Solve the following pair of linear equations by the elimination method and find the value of $p+q$ rounded off to the nearest integer.
$4p=3q+5; \, \, p-q = \dfrac{7}{6}$

$2x-y-3=0; \, \, 4x-y-5=0$

$\displaystyle \frac{3}{2}x + \frac{2}{3}y = 13 \frac{1}{3}; \, \, \frac{2}{3}x + \frac{3}{2}y = 8 \frac{1}{3}$

$3x-2y = 4; \, \, 6x+7y=19$

Solve for $x$ & $y$

$ax+by=a-b$
$bx-ay=a+b$

Find the fraction which becomes to $\dfrac23$ when the numerator is increasing by $2$ and equal to $\dfrac47$ when the denominator is increased by $4$

Solve the following pair of linear equations in two variables:
$3=2x+y$ and $9 = 4x - y$

Solve for $x$ and $y:x-4y-14=0$ , $5x-y-13=0$

Solve using elimination method :
$7x + 2y = 19$
$x - y = 4$

Solve the following system of equation for x and y.
$\displaystyle \frac{5}{x-1}+\frac{1}{y-2}=2,\frac{6}{x-1}-\frac{3}{y-2}=1$

Solve : $\dfrac{x}{a}+\dfrac{y}{b}=2$
$ax-by=a^{2}-b^{2}$

Solve the following systems of linear equations

$i)2x - y = 17$

$ii)3x + 5y = 6$

Solve for $x$ and $y$ :
$2x-y+3=0$ ,
$3x-7y+10=0$ .

Solve for $x$ and $y$ :
$3x-5y-19=0$ ,
$-7x+3y+1=0$ .

Solve for $x$ and $y$ :
$x/3+y/4=11$ ,
$5x/6-y/3=-7$

Solve for $x$ and $y$ :
$2x+3y=0$ ,
$3x+4y=5$ .

Solve the following pair of linear equations by the elimination method :
3x - 5y - 4 = 0 and 9x = 2y + 7

Solve the following pair of linear equations by the elimination method the substitution method :
3x + 4 y = 10 and 2x - 2y = 2

Solve the following pair of linear equations by the elimination method the substitution method :
$\dfrac{x}{2} + \dfrac{2y}{3} = -1 and x - \dfrac{y}{3} = 3$

Solve the following pair of linear equations by suitable method
x + y = 5 and 2x - 3y = 4

Solve the following pair of linear equations
$( a - b) x + ( a + b) y = a^{2} - 2ab - b^{2} \rightarrow (1)$
$( a + b) (x + y) = a^{2} + b^{2}$

Solve the following pair of the equations by reducing them to a pair of linear equations:
$6x + 3y = 6xy$
$2x + 4y = 5xy$

Solve the following pair of linear equations
$px + qy = p - q$
$qx - py = p + q$

Solve the simultaneous equation and find the value of y
$\displaystyle \frac{x+y-8}{2} = \frac{x+2y-14}{3} = \frac{3x+y-12}{11}$