Solve the two equations using method of eliminationa1x+

SOLUTION

Given two equations are
$a_{1}x+b_{1}y+c_1=0$ 
and $a_{2}x+b_{2}y+c_2=0$

On multiplying the first equation by $b_2$ and second equation by $b_1$, we get

$b_2$$a_{1}x$ + $b_2$$b_{1}y$ + $b_2$$c_1$ = 0

$b_1$$a_{2}x$ + $b_1$$b_{2}y$ + $b_1$$c_2$ = 0

On subtracting these two equations, we get

$(b_2{a}_1–b_1{a}_2)x+(b_2{b}_1–b_1{b}_2)y+(b_2{c}_1–b_1{c}_2)=0$

i.e., $(b_2$$a_1$ – $b_1$$a_2)x$ = $b_1$$c_2$ – $b_2$$c_1$

Therefore $x=\frac{(b_1{c}_2-b_2{c}_1)}{(a_1{b}_2-a_2{b}_1)}$ ,

On substituting the value of x in equation (1), we get  

$a_1\left(\frac{b_1{c}_2-b_2{c}_1}{a_1{b}_2-a_2{b}_1}\right)+b_{1}y+c_1=0$ 

$\Rightarrow b_{1}y=-c_1-a_1\left(\frac{b_1{c}_2-b_2{c}_1}{a_1{b}_2-a_2{b}_1}\right)$

$\Rightarrow b_{1}y=\frac{-c_1(a_1{b}_2-a_2{b}_1)-a_1(b_1{c}_2-b_2{c}_1)}{a_1{b}_2-a_2{b}_1}$

$\Rightarrow y=\frac{b_1\times (c_1{a}_2-c_2{a}_1)}{b_1\times (a_1{b}_2-a_2{b}_1)}$

$\Rightarrow y=\frac{(c_1{a}_2-c_2{a}_1)}{(a_1{b}_2-a_2{b}_1)}$