Answer
The solutions are $(1+\sqrt{15},7+2\sqrt{15})$ and $(1-\sqrt{15},7-2\sqrt{15})$.
Work Step by Step
The given system is
$y=x^2-9$ ...... (1)
$y=2x+5$ ...... (2)
Subtract equation (2) from equation (1).
$\Rightarrow y-y=x^2-9-(2x+5)$
$\Rightarrow y-y=x^2-9-2x-5$
Simplify.
$\Rightarrow 0=x^2-2x-14$
Add $15$ to each side.
$\Rightarrow 0+15=x^2-2x-14+15$
Simplify.
$\Rightarrow 15=x^2-2x+1$
Write the right side terms as square of the polynomial.
$\Rightarrow 15=(x-1)^2$
Take square root on each side.
$\Rightarrow \pm\sqrt{15}=x-1$
Add $a$ to each side.
$\Rightarrow 1\pm\sqrt{15}=x-1+1$
Simplify.
$\Rightarrow 1\pm\sqrt{15}=x$
Substitute $1+\sqrt{15}$ for $x$ in equation (2).
$\Rightarrow y=2(1+\sqrt{15})+5$
Simplify.
$\Rightarrow y=2+2\sqrt{15}+5$
$\Rightarrow y=7+2\sqrt{15}$
Substitute $1-\sqrt{15}$ for $x$ in equation (2).
$\Rightarrow y=2(1-\sqrt{15})+5$
Simplify.
$\Rightarrow y=2-2\sqrt{15}+5$
$\Rightarrow y=7-2\sqrt{15}$
Hence, the solutions are $(1+\sqrt{15},7+2\sqrt{15})$ and $(1-\sqrt{15},7-2\sqrt{15})$.
