Question

Solve:

$x+y=a+b$ and
$ax-by=a^2-b^2$.

Answer 1

The given system of equations may be written as

$x+y-(a+b)=0$

$ax-by-(a^2-b^2)=0$

By cross-multiplication, we get

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

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

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

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

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

Hence, the solution of the given system of equations is $x=a,y=b$.