Answer
$(\pm2,-3),(\pm3,2)$
Work Step by Step
Step 1. Add up the two equations to get
$x^2+y^2+y=x^2-7+13$
or
$y^2+y-6=0$
Step 2. Factor and solve the quadratic:
$(y+3)(y-2)=0$
which gives
$y=-3, 2$
Step 3. For $y=-3$, using the first equation, we have
$x^2=-3+7=4$
Thus
$x=\pm2$
Step 4. For $y=2$, use the first equation. We have
$x^2=2+7=9$
Thus
$x=\pm3$
Step 5. The solutions to the system are
$(\pm2,-3),(\pm3,2)$
