Answer
$\{ (2,0),(-2,0)\}$
Work Step by Step
$x^{2}+2y^{2} =4$ Equation $(1)$
$x^{2}-y^{2} =4$ Equation $(2)$
Multiply Equation $(2)$ by $2$ then add with Equation $(1)$
$2(x^{2}-y^{2}) +x^{2}+2y^{2} =2(4)+4$
$2x^{2}-2y^{2} +x^{2}+2y^{2} =8+4$
$3x^{2}=12$
$x^{2}=4$
$x=±2$
$x= 2$ or $x=-2$
Substitute $x$ values in Equation $(2)$
Let $x= 2$
$x^{2}-y^{2} =4$
$2^{2}-y^{2} =4$
$4-y^{2} =4$
$y^{2} =4-4$
$y^{2} =0$
$y =0$
Let $x= -2$
$x^{2}-y^{2} =4$
$(-2)^{2}-y^{2} =4$
$4-y^{2} =4$
$y^{2} =4-4$
$y^{2} =0$
$y =0$
$(2,0),(-2,0)$ satisfy both equations. Solution set is $\{ (2,0),(-2,0)\}$
