Answer
The solutions of the system are $(3,-16)$ and $(-13,64)$.
Work Step by Step
$y+1=-5x \rightarrow y=-5x-1$
$y=x^2+5x-40$
Substitute $-5x-1$ for y
$-5x-1=x^2+5x-40$
$x^2+10x-39=0$
$(x-3)(x+13)=0$
$x-3=0$ or $x+13=0$
$x=3$ or $x=-13$
Find corresponding y-values. Use either original equation
$ y=-5x-1$
$=-5(3)-1$ or $ y=-5(-13)-1$
$y=-16$ or $y=64$
The solutions of the system are $(3,-16)$ and $(-13,64)$.
