Answer
The solutions of the system are $(2,8)$ and $(-\frac{1}{2},\frac{1}{2})$.
Work Step by Step
$3x-y=-2 \rightarrow y=3x+2$
$y=2x^2$
Substitute $3x+2$ for y
$3x+2=2x^2$
$2x^2-3x-2=0$
$(x-2)(2x+1)=0$
$x-2=0$ or $2x+1=0$
$x=2$ or $x=-\frac{1}{2}$
Find corresponding y-values. Use either original equation.
$y=3x+2$
$y=3(2)+2$ or $y=3(-\frac{1}{2})+2$
$y=8$ or $y=\frac{1}{2}$
The solutions of the system are $(2,8)$ and $(-\frac{1}{2},\frac{1}{2})$.
