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ax + by = 1

bx + ay = (a + b)²/ a²+ b²- 1

Solve for x and y.

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

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