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Solve for 'x' and 'y' by using method of substitution.

bx + ay = a + b
$a x [\frac{1}{a - b} - \frac{1}{a + b}] + b y [\frac{1}{b - a} - \frac{1}{b + a}] = 2$

$T h e g i v e n s y s t e m o f e q u a t i o n b e a s, b x + a y = a + b \Rightarrow b x = a + b - a y \Rightarrow x = \frac{a + b - a y}{b} ....... (1) A n d, a x (\frac{1}{a - b} - \frac{1}{a + b}) + b y (\frac{1}{b - a} - \frac{1}{b + a}) = 2 \Rightarrow a x (\frac{a + b - a + b}{a^{2} - b^{2}}) + b y (\frac{b + a - b + a}{b^{2} - a^{2}}) = 2 \Rightarrow a x (\frac{2 b}{a^{2} - b^{2}}) + b y (\frac{2 a}{b^{2} - a^{2}}) = 2 \Rightarrow a x (\frac{b}{a^{2} - b^{2}}) - b y (\frac{a}{a^{2} - b^{2}}) = 1 \Rightarrow x - y = \frac{a^{2} - b^{2}}{a b} ..... (2) S u b s t i t u t e v a l u e o f x f r o m e q u a t i o n (1) i n e q u a t i o n (2), w e h a v e, \frac{a + b - a y}{b} - y = \frac{a^{2} - b^{2}}{a b} \Rightarrow \frac{a + b}{b} - \frac{a}{b} y - y = \frac{a^{2} - b^{2}}{a b} \Rightarrow y (\frac{b + a}{b}) = \frac{a^{2} - b^{2}}{a b} - \frac{a + b}{b} \Rightarrow y (\frac{b + a}{b}) = \frac{a^{2} b - b^{3} - a^{2} b - a b^{2}}{a b^{2}} \Rightarrow y (\frac{b + a}{b}) = \frac{{- b}^{3} - a b^{2}}{a b^{2}} \Rightarrow y (\frac{b + a}{b}) = \frac{{- b}^{2} (b + a)}{a b^{2}} \Rightarrow y = - \frac{b}{a} S u b s t i t u t e v a l u e o f y i n e q u a t i o n (2), w e h a v e, x - \frac{- b}{a} = \frac{a^{2} - b^{2}}{a b} \Rightarrow x + \frac{b}{a} = \frac{a^{2} - b^{2}}{a b} \Rightarrow x = \frac{a^{2} - b^{2}}{a b} - \frac{b}{a} \Rightarrow x = \frac{a^{3} - a b^{2} - a b^{2}}{a^{2} b} = \frac{a}{b} H e n c e, x = \frac{a}{b} a n d y = \frac{b}{a}$

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