Chapter 9 Quadratic Relations and Conic Sections - 9.7 Solve Quadratic Systems - 9.7 Exercises - Skill Practice - Page 661: 18

$(7,-22),(-1,2)$

Substituting the second equation ($y=-1-3x$) into the first one we get: $x^2+6x+4(-1-3x)-3=0\\x^2-6x-7=0\\(x-7)(x+1)=0$ Thus $x=7$ or $x=-1$. If $x=7$, then $y=-22$, and if $x=-1$, then $y=2$. Thus the solutions are: $(7,-22),(-1,2)$