Question

For the given pair of linear equations $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0,$ the solution $(x,y)$ is

• A. ((b₁c₂-b₂c₁)/(a₁b₂-a₂b₁), (c₁a₂-c₂a₁)/(a₁b₂-a₂b₁))
• B. ((b₂c₂-b₁c₁)/(a₁b₂-a₂b₁), (c₂a₂-c₁a₁)/(a₁b₂-a₂b₁))
• C. ((b₂c₁-b₁c₂)/(a₁b₂-a₂b₁), (c₁a₁-c₂a₂)/(a₁b₂-a₂b₁))
• D. None of these

Answer 1

The correct option is A

((b₁c₂-b₂c₁)/(a₁b₂-a₂b₁), (c₁a₂-c₂a₁)/(a₁b₂-a₂b₁))

Let us use Elimination method to solve the given pair of equations

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

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

Multiply first equation by $b_2$ and second equation by $b_1$, to get

$b_2{a}_{1}x+b_2{b}_{1}y+b_2{c}_1=0$
$b_1{a}_{2}x+b_1{b}_{2}y+b_1{c}_2=0$

Subtracting these two equations

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

Substitute this x in any of the given equation we get

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