Answer
$\{\ (3,2),\ (3, -2),\ (-3,2),\ (-3, -2) \ \}$
Work Step by Step
1. Eliminating:
Adding the equations eliminates the $(y^{2})$ terms
$\left\{\begin{array}{l}
x^{2}+y^{2}=13\\
x^{2}-y^{2}=5
\end{array}\right.$
$2x^{2}=18$
2. Solving:
$2x^{2}=18\qquad/\div 2$
$x^{2}=9$
$x=\pm 3$
3. Back-substituting:
$\left[\begin{array}{lll}
x=3 & or & x=-3\\
(3)^{2} +y^{2}=13 & & (-3)^{2}+y^{2}=13\\
y^{2}=4 & & y^{2}=4\\
y=\pm 2 & & y=\pm 2
\end{array}\right]$
The solution set is
$\{\ (3,2),\ (3, -2),\ (-3,2),\ (-3, -2) \ \}$
