Chapter 9 Quadratic Relations and Conic Sections - 9.7 Solve Quadratic Systems - 9.7 Exercises - Skill Practice - Page 661: 9

$(-2,2),(3,-3)$

Substituting the second equation into the first one we get: $(-x)^2-x-6=0\\x^2-x-6=0\\(x+2)(x-3)=0$ Thus $x=-2$ or $x=3$. If $x=-2$, then $y=2$, and if $x=3$, then $y=-3$. Thus the solutions are: $(-2,2),(3,-3)$