Answer
The solutions are $(0,-5)$ and $(-3,-8)$.
Work Step by Step
The given system is
$y=x-5$ ...... (1)
$y=x^2+4x-5$ ...... (2)
Substitute $x-5$ for $y$ in equation (2).
$\Rightarrow x-5=x^2+4x-5$
Add $5-x$ to each side.
$\Rightarrow x-5+5-x=x^2+4x-5+5-x$
Simplify.
$\Rightarrow 0=x^2+3x$
Factor out $x$.
$\Rightarrow 0=x(x+3)$
Use zero product property.
$\Rightarrow x=0$ or $x+3=0$
Solve for $x$.
$\Rightarrow x=0$ or $x=-3$
Substitute $0$ for $x$ in equation (1).
$\Rightarrow y=0-5$
Simplify.
$\Rightarrow y=-5$
Substitute $-3$ for $x$ in equation (1).
$\Rightarrow y=-3-5$
Simplify.
$\Rightarrow y=-8$
Hence, the solutions are $(0,-5)$ and $(-3,-8)$.
