Question

Solve for $x$, $ax+by=c$ and

$bx+ay=1+c$.

Answer 1

The given system of equations may be written as
$ax+by-c=0$
$bx+ay-(1+c)=0$

By cross-multiplication, we have
$\frac{x}{b\times -(1+c)-a\times -c}=\frac{-y}{a\times -(1+c)-b\times -c}=\frac{1}{a\times a-b\times b}$

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

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

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

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

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

Hence, the solution of the given system of equations is
$x=\frac{c}{a+b}-\frac{b}{a^2-b^2},y=\frac{c}{a+b}+\frac{a}{a^2-b^2}$.