## Precalculus (6th Edition) Blitzer

$(\pm2,-3),(\pm3,2)$
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 $(\pm2,-3),(\pm3,2)$