Answer
$\{\}$ or No solution.
Work Step by Step
$x^{2}+2y^{2} = 2$ Equation $(1)$
$x-y = 2$ Equation $(2)$
From Equation $(2)$,
$x = 2 + y$
$x = y + 2$
Substitute $x=y+2$ in Equation $(1)$
$x^{2}+2y^{2} = 2$
$(y+2)^{2}+2y^{2} = 2$
Using $(a+b)^{2} = a^{2}+2ab+b^{2}$
$(y+2)^{2}=y^{2} + 4y +4$
$y^{2} + 4y +4 +2y^{2} = 2$
$3y^{2} + 4y +4 -2 = 0$
$3y^{2} + 4y +2 = 0$
Using quadratic formula, $a=3, b = 4, c=2$
$y = \frac{-b±\sqrt (b^{2}-4ac)}{2a}$
$y = \frac{-4±\sqrt (4^{2}-4(3)(2))}{2(3)}$
$y = \frac{-4±\sqrt (16-24)}{6}$
$y = \frac{-4±\sqrt (-8)}{6}$
Since $\sqrt (-8)$ is not a real number, there is no real solution.
Solution set is empty.
