Chapter 9 Quadratic Relations and Conic Sections - 9.7 Solve Quadratic Systems - 9.7 Exercises - Quiz for Lessons 9.6-9.7 - Page 664: 8

$(-8,1),(-24,3)$

Subtracting the second equation from the first one we get: $2y^2-8y+6=0\\y^2-4y+3=0\\(y-1)(y-3)=0$ Thus $y=1$ or $y=3$ $x=-8y$, thus if $y=1$, then $x=-8$, and if $y=3$, then $x=-24$. Thus the solutions are: $(-8,1),(-24,3)$