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Question Number 160302 by cortano last updated on 27/Nov/21

	Answered by Rasheed.Sindhi last updated on 27/Nov/21
	${ ((8x^2 −8y^2 =24)),((3x^2 +3xy+6y^2 =24)) :}⇒5x^2 −3xy−14y^2 =0 ⇒5x^2 −10xy+7xy−14y^2 =0 5x(x−2y)+7y(x−2y)=0 (x−2y)(5x+7y)=0 x=2y ∣ x=−((7y)/5) x=2y x^2 −y^2 =3⇒4y^2 −y^2 =3⇒y=±1 ⇒x=2(±1)=±2 (x,y)=(1,2),(−1,−2) x=−((7y)/5) (−((7y)/5))^2 −y^2 =3⇒49y^2 −25y^2 =75 ⇒24y^2 =75⇒y^2 =((25)/8)⇒y=±(5/(2(√2))) ⇒x=−(7/5)(±(5/(2(√2))))=∓(7/( 2(√2))) (x,y)=((7/( 2(√2))),−(5/(2(√2)))),(−(7/( 2(√2))),(5/(2(√2)))) (x,y)=(1,2),(−1,−2),((7/( 2(√2))),−(5/(2(√2)))),(−(7/( 2(√2))),(5/(2(√2))))$
	${\begin{cases} 8 x^{2} - 8 y^{2} = 24 \\ 3 x^{2} + 3 x y + 6 y^{2} = 24 \end{cases} \Rightarrow 5 x^{2} - 3 x y - 14 y^{2} = 0$ $\Rightarrow 5 x^{2} - 10 x y + 7 x y - 14 y^{2} = 0$ $5 x (x - 2 y) + 7 y (x - 2 y) = 0$ $(x - 2 y) (5 x + 7 y) = 0$ $x = 2 y ∣ x = - \frac{7 y}{5}$ $\underset{―}{x = 2 y}$ $x^{2} - y^{2} = 3 \Rightarrow 4 y^{2} - y^{2} = 3 \Rightarrow y = \pm 1$ $\Rightarrow x = 2 (\pm 1) = \pm 2$ $(x, y) = (1, 2), (- 1, - 2)$ $x = - \frac{7 y}{5}$ ${(- \frac{7 y}{5})}^{2} - y^{2} = 3 \Rightarrow 49 y^{2} - 25 y^{2} = 75$ $\Rightarrow 24 y^{2} = 75 \Rightarrow y^{2} = \frac{25}{8} \Rightarrow y = \pm \frac{5}{2 \sqrt{2}}$ $\Rightarrow x = - \frac{7}{5} (\pm \frac{5}{2 \sqrt{2}}) = \mp \frac{7}{2 \sqrt{2}}$ $(x, y) = (\frac{7}{2 \sqrt{2}}, - \frac{5}{2 \sqrt{2}}), (- \frac{7}{2 \sqrt{2}}, \frac{5}{2 \sqrt{2}})$ $(x, y) = (1, 2), (- 1, - 2), (\frac{7}{2 \sqrt{2}}, - \frac{5}{2 \sqrt{2}}), (- \frac{7}{2 \sqrt{2}}, \frac{5}{2 \sqrt{2}})$
	Commented by cortano last updated on 27/Nov/21

	$yes nice$
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