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Question Number 187476 by mnjuly1970 last updated on 17/Feb/23

	Answered by anurup last updated on 18/Feb/23

	$If 2 x^{2} + 3 x y - 2 y^{2} - 5 (2 x - y) = 0$ $x^{2} - 2 x y - 3 y^{2} + 15 = 0$ $find x, y .$ $Solution :$ $2 x^{2} + 3 x y - 2 y^{2} - 5 (2 x - y) = 0$ $2 x^{2} + 4 x y - x y - 2 y^{2} - 5 (2 x - y) = 0$ $\Rightarrow 2 x (x + 2 y) - y (x + 2 y) - 5 (2 x - y) = 0$ $\Rightarrow (x + 2 y) (2 x - y) - 5 (2 x - y) = 0$ $\Rightarrow (2 x - y) (x + 2 y - 5) = 0$ $\Rightarrow y = 2 x or x + 2 y = 5$ $x^{2} - 2 x y - 3 y^{2} + 15 = 0$ $\Rightarrow (x^{2} - 2 x y + y^{2}) - 4 y^{2} + 15 = 0$ $\Rightarrow {(x - y)}^{2} - {(2 y)}^{2} + 15 = 0$ $\Rightarrow (x + y) (x - 3 y) + 15 = 0$ $If y = 2 x then$ $3 x (- 5 x) + 15 = 0$ $\Rightarrow x^{2} = 1$ $\Rightarrow x = \pm 1 and y = \pm 2$ $If x = 5 - 2 y then$ $(5 - y) (5 - 5 y) + 15 = 0$ $\Rightarrow (y - 5) (y - 1) + 3 = 0$ $\Rightarrow y^{2} - 6 y + 8 = 0$ $\Rightarrow (y - 4) (y - 2) = 0$ $\Rightarrow y = 4, y = 2 and x = - 3, x = 1$ $Solutions are (\pm 1, \pm 2), (- 3, 4)$
	Commented by anurup last updated on 17/Feb/23

	${it}^{'} s quite easy$
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