Answer
$(2,2),(-2,-2),(\sqrt 2,2\sqrt 2),(-\sqrt 2,-2\sqrt 2)$
Work Step by Step
1. From the 2nd equation, we can get $y=\frac{4}{x}$
2. Plugin the 1st equation to get $2x^2-x(\frac{4}{x})+(\frac{4}{x})^2=8$ or $x^2-2+\frac{8}{x^2}=4$
3. Multiply $x^2$ to get $x^4-6x^2+8=0$ or $(x^2-2)(x^2-4)=0$, thus $x=\pm2, \pm\sqrt 2$
4. Use the above with the 2nd equation to get the solutions $(2,2),(-2,-2),(\sqrt 2,2\sqrt 2),(-\sqrt 2,-2\sqrt 2)$
