Solve the pair of linear equations by using cross multiplication method
(iii) $(a - b) x + (a + b) y = 2 a^{2} - 2 a b^{2}, (a + b) (x + y) = 4 a b$

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Solution

$(a - b) x + (a + b) y = 2 a^{2} - 2 a b^{2}$
$(a + b) x + (a + b) y = 4 a b$
$\frac{x}{4 a b (a + b) - (a + b) (2 a^{2} - 2 a b^{2})} = \frac{y}{(a + b) (2 a^{2} - 2 a b^{2}) - 4 a b (a - b)} = \frac{1}{a^{2} - b^{2} - a^{2} - b^{2} - 2 a b}$
$\Rightarrow \frac{x}{(a + b) (4 a b - 2 a^{2} + 2 a b^{2})} = \frac{1}{- 2 b^{2} - 2 a b}$ and $\frac{y}{2 a (a + b) (a - b^{2}) - 4 a b (a - b)} = \frac{1}{- 2 b^{2} - 2 a b}$
$\Rightarrow x = \frac{(a + b) (4 a b - 2 a^{2} + 2 a b^{2})}{- 2 b (b + a)}$ and $y = \frac{2 a (a + b) (a - b^{2}) - 4 a b (a - b)}{- 2 b (b + a)}$
$\Rightarrow x = \frac{4 a b - 2 a^{2} + 2 a b^{2}}{- 2 b}$ and $y = \frac{a (a + b) (a - b^{2}) - 2 a b (a - b)}{- b (b + a)}$
$∴ x = \frac{- 2 a b + a^{2} - a b^{2}}{b}$ and $y = \frac{a (a + b) (a - b^{2}) - 2 a b (a - b)}{- b (b + a)}$

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Similar questions

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

(a + b) (x + y) = 4 a b

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Solve the equation by substitution method :

(a+b)x + (a+b)y = 2a² -2b²

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\frac{x}{a} + \frac{y}{b} = a + b, \frac{x}{a^{2}} + \frac{y}{b^{2}} = 2

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