Chapter 9 Quadratic Relations and Conic Sections - 9.7 Solve Quadratic Systems - Guided Practice for Examples 1 and 2 - Page 659: 5

$(3,-4)$ and $(-3,2)$

Substituting the second equation into the first one we get: $(-x-1)^2-2x-10=0\\x^2+2x+1-2x-10=0\\x^2-9=0\\x^2=9\\x=\pm3$ When $x=3$, then $y=-3-1=-4$, and when $x=-3$, then $y=-(-3)-1=2$. Thus the solutions are $(3,-4)$ and $(-3,2)$.