Question

Solve the following pair of linear equation
$ax+by=c$
$bx+ay=1+c$

Answer 1

Given equations are

$ax+by=c$

$bx+ay=1+c$

By cross multiplication, we get

$\frac{x}{-b(1+c)+ac}=\frac{-y}{-a(1+c)+bc}=\frac{1}{a^2-b^2}$

$\Rightarrow \frac{x}{ac-b-bc}=\frac{-y}{-bc+ac-a}=\frac{1}{a^2-b^2}$

$\Rightarrow \frac{x}{c(a-b)-b}=\frac{y}{c(a-b)+a}=\frac{1}{(a+b)(a-b)}$

$\Rightarrow x=\frac{c(a-b)-b}{(a+b)(a-b)}$ and $y=\frac{c(a-b)+a}{(a+b)(a-b)}$

$\Rightarrow x=\frac{c}{a+b}-\frac{b}{(a+b)(a-b)}=\frac{c}{(a+b)}-\frac{b}{a^2-b^2}$

and $y=\frac{c}{a+b}+\frac{a}{(a+b)(a-b)}=\frac{c}{a+b}+\frac{a}{a^2+b^2}$.