Solve the following pair of equations by cross multiplication rule.
$x + y = a + b, a x - b y = a^{2} - b^{2}$

x = 1, y = a b

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x = a + b, y = b

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x = a, y = b

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x = b, y = 1

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Solution

The correct option is C $x = a, y = b$
The equations are
$\Rightarrow x + y - a - b = 0$

$\Rightarrow a x - b y - a^{2} + b^{2} = 0$
$B y c r o s s m u l t i p l i c a t i o n m e t h o d$
$\Rightarrow \frac{x}{- (a^{2} - b^{2}) - [- (a + b) (- b)]} = \frac{y}{- (a + b) a - [- (a^{2} - b^{2})]} = \frac{1}{- b - a}$
$\Rightarrow \frac{x}{- a^{2} - b^{2} - (a b + b^{2})} = \frac{y}{a^{2} - a b + (a^{2} - b^{2})} = \frac{1}{- b - a}$

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

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