Answer
The solutions are $(-3,-2)$ and $(1,6)$.
Work Step by Step
The given system is
$y=-x^2+7$ ...... (1)
$y=2x+4$ ...... (2)
Substitute $2x+4$ for $y$ in equation (1).
$\Rightarrow 2x+4=-x^2+7$
Add $x^2-7$ to each side.
$\Rightarrow 2x+4+x^2-7=-x^2+7+x^2-7$
Simplify.
$\Rightarrow x^2+2x-3=0$
Factor the polynomial.
$\Rightarrow (x+3)(x-1)$
Use zero product property.
$\Rightarrow x+3=0$ or $x-1=0$
Solve for $x$.
$\Rightarrow x=-3$ or $x=1$
Substitute $-3$ for $x$ in equation (2).
$\Rightarrow y=2(-3)+4$
Simplify.
$\Rightarrow y=-6+4$
$\Rightarrow y=-2$
Substitute $1$ for $x$ in equation (1).
$\Rightarrow y=2(1)+4$
Simplify.
$\Rightarrow y=2+4$
$\Rightarrow y=6$
Hence, the solutions are $(-3,-2)$ and $(1,6)$.
