Answer
The solutions of the system are $(3,13)$ and $(-4,6)$.
Work Step by Step
$ y=x+10$
$y=x^2+2x-2$
Substitute $x+10$ for y
$x+10=x^2+2x-2$
$x^2+x-12=0$
$(x-3)(x+4)=0$
$x-3=0$ or $x+4=0$
$x=3$ or $x=-4$
Find corresponding y-values. Use either original equation
$ y=x+10$
$y=3+10$ or $ y=(-4)+10$
$y=13$ or $y=6$
The solutions of the system are $(3,13)$ and $(-4,6)$.
