Answer
$\{\ (1,0),\ (-1,0)\ \}$
Work Step by Step
1. Eliminating:
Adding the equations eliminates the $(y^{2})$ terms
$\left\{\begin{array}{l}
4x^{2}-y^{2}=4\\
4x^{2}+y^{2}=4
\end{array}\right.$
$8x^{2}=8$
2. Solving:
$ 8x^{2}=8\qquad/\div$8
$x^{2}=1$
$x=\pm 1$
3. Back-substituting:
$\left[\begin{array}{lll}
x=1 & or & x=-1\\
4(1)^{2}+y^{2}=4 & & 4 (-1)^{2}+y^{2}=4\\
y^{2}=0 & & y^{2}=0\\
y=0 & & y=0
\end{array}\right]$
The solution set is
$\{\ (1,0),\ (-1,0)\ \}$
