Answer
$(0,0),(3,3),(-3,3)$
Work Step by Step
1. Take the difference between the two equations to get $y^2=3y$, thus $y=0,3$
2. For $y=0$, use the 2nd equation to get $x=0$
3. For $y=3$, use the 2nd equation to get $x^2=9$, thus $x=\pm3$
4. We have the solution(s) $(0,0),(3,3),(-3,3)$
