Answer
$(\sqrt 3,\sqrt 3),(-\sqrt 3,-\sqrt 3),(2,1),(-2,-1)$
Work Step by Step
1. Factor the first equation to get $(x-y)(x-2y)=0$
2. For $y=x$, use the 2nd equation, we have:
$x^2+x^2=6$ or $x^2=3$, thus $x=\pm\sqrt 3$ and $y=x=\pm\sqrt 3$
3. For $y=x/2$, use the 2nd equation, we have:
$x^2+x^2/2=6$ or $x^2=4$, thus $x=\pm2$ and $y=x/2=\pm1$
4. Combine the above results to get the solutions:
$(\sqrt 3,\sqrt 3),(-\sqrt 3,-\sqrt 3),(2,1),(-2,-1)$
