Answer
$(2,\sqrt 2), (2,-\sqrt 2), (-2,\sqrt 2), (-2,-\sqrt 2)$
Work Step by Step
1. Multiply -3 to the 1st equation and add to the 2nd, we have $2/y^2=1$, thus $y^2=2$ and $y=\pm\sqrt 2$
2. Use the 1st equation, $2/(x)^2-3/2+1=0$ or $x^2=4$, thus $x=\pm2$
3. The solutions $(2,\sqrt 2), (2,-\sqrt 2), (-2,\sqrt 2), (-2,-\sqrt 2)$
