Answer
$\{\ (3,2),\ (3,-2),\ (-3,2),\ (-3,-2)\ \}$
Work Step by Step
1. Eliminating:
eliminate the $(y^{2})$ terms:
$\left\{\begin{array}{lll}
x^{2}-4y^{2}=-7 & , & \\
3x^{2}+y^{2}=31 & , & /\times 4, add
\end{array}\right.$
$(1+12)x^{2}=-7+124$
$13x^{2}=117$
2. Solving:
$13x^{2}=117\qquad/\div 13$
$x^{2}=9$
$x=\pm 3$
3. Back-substituting:
$\left[\begin{array}{lll}
x=3 & or & x=-3\\
3(9)+y^{2}=31 & & 3(9)+y^{2}=31 \\
y^{2}=4 & & y^{2}=4\\
y=\pm 2 & & y=\pm 2
\end{array}\right]$
The solution set is
$\{\ (3,2),\ (3,-2),\ (-3,2),\ (-3,-2)\ \}$
