Answer
The solutions of the system are $(-10, 130)$ and $(-7,100)$.
Work Step by Step
$y+10x=30 \rightarrow y=30-10x$
$y=x^2+7x+100$
Substitute $30-10x$ for y
$30-10x=x^2+7x+100$
$x^2+17x+70=0$
$(x+10)(x+7)=0$
$x+10=0$ or $x+7=0$
$x=-10$ or $x=-7$
Find corresponding y-values. Use either original equation
$y=30-10x$
$y=30-10(-10)$ or $y=30-10(-7)$
$y=130$ or $y=100$
The solutions of the system are $(-10, 130)$ and $(-7,100)$.
