Answer
$(\sqrt 3,\sqrt 5),(-\sqrt 3,\sqrt 5),(\sqrt 3,-\sqrt 5),(-\sqrt 3,-\sqrt 5)$
Work Step by Step
$4x^{2}-2y^{2} =2$ Equation $(1)$
$-x^{2}+y^{2} =2$ Equation $(2)$
Multiply Equation $(2)$ by $4$ and add with Equation $(1)$
$4(-x^{2}+y^{2})+4x^{2}-2y^{2} =4(2)+2$
$-4x^{2}+4y^{2}+4x^{2}-2y^{2} =8+2$
$2y^{2} =10$
$y^{2} =5$
$y=±\sqrt 5$
$y=\sqrt 5$ or $y=-\sqrt 5$
Substitute $y$ values in Equation $(1)$
$y=\sqrt 5$
$4x^{2}-2y^{2} =2$
$4x^{2}-2(\sqrt 5)^{2} =2$
$4x^{2}-2(5) =2$
$4x^{2}-10 =2$
$4x^{2}=2+10$
$4x^{2}=12$
$x^{2}=3$
$x=±\sqrt 3$
$y=-\sqrt 5$
$4x^{2}-2y^{2} =2$
$4x^{2}-2(-\sqrt 5)^{2} =2$
$4x^{2}-2(5) =2$
$4x^{2}-10 =2$
$4x^{2}=2+10$
$4x^{2}=12$
$x^{2}=3$
$x=±\sqrt 3$
$(\sqrt 3,\sqrt 5),(-\sqrt 3,\sqrt 5),(\sqrt 3,-\sqrt 5),(-\sqrt 3,-\sqrt 5)$
satisfy both the equations. So the solution set is $\{(\sqrt 3,\sqrt 5),(-\sqrt 3,\sqrt 5),(\sqrt 3,-\sqrt 5),(-\sqrt 3,-\sqrt 5)\}$
