Answer
$\{\}$ or No solution
Work Step by Step
$x^{2}+2y^{2} = 2$ Equation $(1)$
$x^{2}-2y^{2} = 6$ Equation $(2)$
Adding Equation $(1)$ and Equation $(2)$
$x^{2}+2y^{2} + x^{2}- 2y^{2} = 2 + 6 $
$2x^{2}= 8$
$x^{2}= 4$
$x = ±2$
$x=2$ or $x=-2$
Substitute $ x$ values in Equation $(1)$
Let $x = 2$
$x^{2}+2y^{2} = 2$
$2^{2}+2y^{2} = 2$
$4+2y^{2} = 2$
$2y^{2} = 2-4$
$2y^{2} = -2$
$y^{2} =-1$
$y = ±\sqrt (-1)$
Since $\sqrt (-1)$ is not a real number, there is no real solution.
