Solve the linear equations by cross multiplication method :

$(a + b) x + (a - b) y = 3 a - b (2 a^{2} + b^{2}) x - (a^{2} - b) y = b^{2} + 2 b$

x=1 and y=3

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x=2 and y= 3

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x=1 and y= 2

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x=1 and y= 4

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Solution

The correct option is C
x=1 and y= 2
$(a + b) x + (a - b) y = 3 a - b (2 a^{2} + b^{2}) x - (a^{2} - b) y = b^{2} + 2 b$

By cross multiplication method :

$\frac{x}{(a - b) (b^{2} + 2 b) - (3 a - b) (- a^{2} + b)} = \frac{y}{(3 a - b) (2 a^{2} + b^{2}) - (b^{2} + 2 b) (a + b)} = \frac{- 1}{- (a + b) (a^{2} - b) - (2 a^{2} + b^{2}) (a - b)} \frac{x}{3 a^{3} - b^{3} + a b^{2} - a^{2} b - b^{2} - a b} = \frac{y}{6 a^{2} - 2 b^{3} + 2 a b^{2} - 2 a b^{2} - 2 a^{2} b - 2 b^{2} - 2 a b} = \frac{- 1}{- (3 a^{3} - b^{3} + a b^{2} - a^{2} b - b^{2} - a b)} \Rightarrow \frac{x}{(3 a^{3} - b^{3} + a b^{2} - a^{2} b - b^{2} - a b)} = \frac{- 1}{- (3 a^{3} - b^{3} + a b^{2} - a^{2} b - b^{2} - a b)} \Rightarrow \frac{y}{2 (3 a^{3} - b^{3} + a b^{2} - a^{2} b - b^{2} - a b)} = \frac{- 1}{- (3 a^{3} - b^{3} + a b^{2} - a^{2} b - b^{2} - a b)} \Rightarrow x = \frac{y}{2} = 1 \Rightarrow x = 1, y = 2$

So, option c is correct.

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