Solve the following system of equations:
$x + y = a - b$
$a x - b y = a^{2} + b^{2}$

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Solution

The given system of equations may be written as
$x + y - (a - b) = 0$
$a x - b y - (a^{2} + b^{2}) = 0$

By cross-multiplication, we have
$\frac{x}{- (a^{2} + b^{2}) - b (a - b)} = \frac{- y}{- (a^{2} + b^{2}) - a \times - (a - b)} = \frac{1}{- b - a}$

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

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

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

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

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