Question

Solve:

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

Answer 1

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

By cross-multiplication, we get

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

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

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

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

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

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