Answer
$\{\ (0,-4),\ (2\sqrt{3},2),\ (-2\sqrt{3},2)\ \}$
Work Step by Step
1. Eliminating:
Eliminate the $( x^{2})$ terms:
$\left\{\begin{array}{lll}
x^{2}-2y=8 & , & / \times(-1)\\
x^{2}+y^{2}=16 & , & / add
\end{array}\right.$
$y^{2}+2y=8$
2. Solving:
$(y+4)(y-2)=0$
$y+4=0$ or $y-2=0$
$y=-4$ or $y=2$
3. Back-substituting into $x^{2}-2y=8$
$\left[\begin{array}{lll}
y=-4 & or & y=2\\
x^{2}-2(-4)=8 & & x^{2}-2(2)=8 \\
x^{2}=8-8 & & x^{2}=8+4\\
x^{2}=0 & & x^{2}=12\\
x=0 & & x=\pm\sqrt{12}\\
& & x=\pm 2\sqrt{3}
\end{array}\right]$
The solution set is
$\{\ (0,-4),\ (2\sqrt{3},2),\ (-2\sqrt{3},2)\ \}$
