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

Select the right options for the values of x and y.

$x = a + b$

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$x = a - b$

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$y = \frac{2 a b}{a - b}$

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$y = \frac{- 2 a b}{a + b}$

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Solution

The correct options are
A
$x = a + b$

D
$y = \frac{- 2 a b}{a + b}$

$(a - b) x + (a + b) y = a^{2} - 2 a b - b^{2}$ ....(1)
$(a + b) (x + y) = a^{2} + b^{2}$
$(a + b) x + (a + b) y = a^{2} + b^{2}$ .....(2)
Subtracting equation (2) from (1), we obtain
$(a - b) x - (a + b) x = (a^{2} - 2 a b - b^{2}) - (a^{2} + b^{2})$
$(a - b - a - b) x = - 2 a b - 2 b^{2}$
$- 2 b x = - 2 b (a + b)$
$x = a + b$
Using equation (1), we obtain
$(a - b) (a + b) + (a + b) y = a^{2} - 2 a b - b^{2}$
$a^{2} - b^{2} + (a + b) y = a^{2} - 2 a b - b^{2}$ [since $(a - b) (a + b) = a^{2} - b^{2}$ ]
$(a + b) y = - 2 a b$
$y = \frac{- 2 a b}{a + b}$

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Similar questions

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

Solve the following pair of linear equation:

(a - b)x + (a + b)y = a² - 2ab - b²

( a + b)(x + y)= a² +b²

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(a−b)x+(a+b)y=a²−2ab−b²

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

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

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