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how to solve (a-b)x + (a+b)y = (a²-2ab-b²) , (a+b) (x+y)=a²+b². by cross multiplication method

$Given (a - b) x + (a + b) y = (a^{2} - 2 a b - b^{2}) (a + b) (x + y) = a^{2} + b^{2} \Rightarrow (a + b) x + (a + b) y = a^{2} + b^{2} By using cross multiplication method we get x y 1 a + b - (a^{2} - 2 a b - b^{2}) a - b a + b a + b - (a^{2} + b^{2}) a + b a + b \Rightarrow \frac{x}{(a + b) (- (a^{2} + b^{2})) + (a + b) (a^{2} - 2 a b - b^{2})} = \frac{y}{- (a^{2} - 2 a b - b^{2}) (a + b) + (a - b) (a^{2} + b^{2})} = \frac{1}{(a - b) (a + b) - (a + b)^{2}} \Rightarrow \frac{x}{(a + b) (- a^{2} - b^{2} + a^{2} - 2 a b - b^{2})} = \frac{y}{- a^{3} - a^{2} b + 2 a^{2} b + 2 a b^{2} + a b^{2} + b^{3} + a^{3} + a b^{2} - a^{2} b - b^{3}} = \frac{1}{a^{2} - b^{2} - a^{2} - 2 a b - b^{2}} \Rightarrow \frac{x}{(a + b) (- 2 b (a + b))} = \frac{y}{4 a b^{2}} = \frac{1}{- 2 b (a + b)} \Rightarrow x = a + b y = \frac{- 2 a b}{a + b}$

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