Obtain the solution of the following pair of equations by cross multiplication method: $a x + b y = 1$ and $b x + a y = \frac{2 a b}{a^{2} + b^{2}}$

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Solution

By cross multiplication method, we get

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

$⟹ \frac{x}{\frac{- 2 a b^{2} + a^{3} + a b^{2}}{a^{2} + b^{2}}} = \frac{y}{\frac{- a^{2} b - b^{3} + 2 a^{2} b}{a^{2} + b^{2}}} = \frac{1}{a^{2} - b^{2}}$

$⟹ \frac{x}{\frac{a^{3} - a b^{2}}{a^{2} + b^{2}}} = \frac{y}{\frac{- b^{3} + a^{2} b}{a^{2} + b^{2}}} = \frac{1}{a^{2} - b^{2}}$

$⟹ \frac{x}{\frac{a^{3} - a b^{2}}{a^{2} + b^{2}}} = \frac{1}{a^{2} - b^{2}}$ and $\frac{y}{\frac{- b^{3} + a^{2} b}{a^{2} + b^{2}}} = \frac{1}{a^{2} - b^{2}}$

$⟹ x = \frac{a^{3} - a b^{2}}{(a^{2} + b^{2}) (a^{2} - b^{2})}$ and $y = \frac{- b^{3} + a^{2} b}{(a^{2} + b^{2}) (a^{2} - b^{2})}$

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

$∴ x = \frac{a}{a^{2} + b^{2}}$ and $y = \frac{b}{a^{2} + b^{2}}$

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