Answer
$(2\sqrt 2,\sqrt 3),(-2\sqrt 2,\sqrt 3),(2\sqrt 2,-\sqrt 3),(-2\sqrt 2,-\sqrt 3),$
Work Step by Step
1. Multiply -2 to the 1st equation and add to the 2nd, we have
$-y^2+3=0$, thus $y=\pm\sqrt 3$
2. Use the 1st equation, $x^2-3(\pm\sqrt 3)^2+1=0$ or $x^2=8$ and $x=\pm2\sqrt 2$
3. The solutions $(2\sqrt 2,\sqrt 3),(-2\sqrt 2,\sqrt 3),(2\sqrt 2,-\sqrt 3),(-2\sqrt 2,-\sqrt 3),$
