Answer
The solutions are $(\sqrt{5},5+\sqrt{5})$ and $(-\sqrt{5},5-\sqrt{5})$.
Work Step by Step
The given system is
$y=x^2+x$ ...... (1)
$y=x+5$ ...... (2)
Subtract equation (2) from equation (1).
$\Rightarrow y-y=x^2+x-(x+5)$
$\Rightarrow y-y=x^2+x-x-5$
Simplify.
$\Rightarrow 0=x^2-5$
Add $5$ to each side.
$\Rightarrow 0+5=x^2-5+5$
Simplify.
$\Rightarrow 5=x^2$
Take square root on each side.
$\Rightarrow \pm\sqrt{5}=x$
Substitute $\sqrt{5}$ for $x$ in equation (2).
$\Rightarrow y=\sqrt{5}+5$
Simplify.
$\Rightarrow y=5+\sqrt{5}$
Substitute $-\sqrt{5}$ for $x$ in equation (2).
$\Rightarrow y=-\sqrt{5}+5$
Simplify.
$\Rightarrow y=5-\sqrt{5}$
Hence, the solutions are $(\sqrt{5},5+\sqrt{5})$ and $(-\sqrt{5},5-\sqrt{5})$.
