Question

(a+b)x + (a-b)y = a^2 - 2ab -b^2 ; (a+b)x + (a+b)y = a^2 +b^2 find x and y

Answer 1

Answer:

answer is 6

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Answer 2

$\large\underline{\sf{Solution-}}$

Given pair of linear equations are

$\rm :\longmapsto\:(a + b)x + (a - b)y = {a}^{2} - 2ab - {b}^{2} - - (1)$

and

$\rm :\longmapsto\:(a + b)x + (a + b)y = {a}^{2} + {b}^{2} - - (2)$

On Subtracting equation (2) from equation (1), we get

$\rm :\longmapsto\: - 2by = - 2ab - 2 {b}^{2}$

$\rm :\longmapsto\: - 2by = - 2b(a + {b})$

$\bf\implies \:y = a + b$

On substituting the value of y, in equation (2), we get

$\rm :\longmapsto\:(a + b)x + (a + b)(a + b) = {a}^{2} + {b}^{2}$

$\rm :\longmapsto\:(a + b)x + {a}^{2} + 2ab + {b}^{2} = {a}^{2} + {b}^{2}$

$\rm :\longmapsto\:(a + b)x + 2ab = 0$

$\rm :\longmapsto\:(a + b)x = - 2ab$

$\bf\implies \:x = - \: \dfrac{2ab}{a + b}$

Verification :-

Consider Line (1) LHS

$\rm :\longmapsto\:(a + b)x + (a - b)y$

On substituting the values of x and y, we get

$\rm \: = \: \: (a + b)\bigg( - \dfrac{2ab}{a + b} \bigg) + (a - b)(a + b)$

$\rm \: = \: \: - 2ab + {a}^{2} - {b}^{2}$

$\rm \: = \: \: {a}^{2} - 2ab - {b}^{2}$

Hence, Verified

Consider line (2) LHS

$\rm :\longmapsto\:(a + b)x + (a + b)y$

On substituting the values of x and y, we get

$\rm \: = \: \: (a + b)\bigg( - \dfrac{2ab}{a + b} \bigg) + (a + b)(a + b)$

$\rm \: = \: \: - 2ab + {a}^{2} + {b}^{2} + 2ab$

$\rm \: = \: \: {a}^{2} + {b}^{2}$

Hence, Verified