Answer
The solutions of the system are $(8, 42)$ and $(-2,2)$.
Work Step by Step
$y=4x+10$
$y=x^2-2x-6$
Substitute $4x+10$ for y:
$4x+10=x^2-2x-6$
$x^2-6x-16=0$
$(x-8)(x+2)=0$
$x-8=0$ or $x+2=0$
$x=8$ or $x=-2$
Find corresponding y-values. Use either original equation $y=4x+10$
$y=4(8)+10$ or $y=4(-2)+10$
$y=42$ or $y=2$
The solutions of the system are $(8, 42)$ and $(-2,2)$.