Answer
$(-2\sqrt 2,-\sqrt 2),(2\sqrt 2,\sqrt 2)$
Work Step by Step
1. Multiply 2 to the 2nd equation and add to the 1st to get $7y^2=14$, thus $y=\pm\sqrt 2$
2. For $y=\sqrt 2$, use the 1st equation to get $2x(\sqrt 2)+(\sqrt 2)^2=10$ or $2\sqrt 2x=8$, thus $x=2\sqrt 2$
3. For $y=-\sqrt 2$, use the 1st equation to get $2x(-\sqrt 2)+(-\sqrt 2)^2=10$ or $-2\sqrt 2x=8$, thus $x=-2\sqrt 2$
4. We have the solution(s) $(-2\sqrt 2,-\sqrt 2),(2\sqrt 2,\sqrt 2)$
