Answer
$\{\ (2,1),\ (2, -1),\ (-2,1),\ (-2, -1)\ \}$
Work Step by Step
1. Eliminating:
Rewrite the equations so the constants are on the RHS.
Eliminate the $(y^{2})$ terms:
$\left\{\begin{array}{lll}
3x^{2}+4y^{2}=16 & , & /\times 3\\
2x^{2}-3y^{2}=5 & , & /\times 4, add
\end{array}\right.$
$(9+8)x^{2}=48+20$
$17x^{2}=68$
2. Solving:
$17x^{2}=68\qquad/\div 17$
$x^{2}=4$
$x=\pm 2$
3. Back-substituting into $3x^{2}+4y^{2}=16$:
$\left[\begin{array}{lll}
x=2 & or & x=-2\\
3(4)+4y^{2}=16 & & 3(4)+4y^{2}=16 \\
4y^{2}=4 & & 4y^{2}=4\\
y^{2}=1 & & y^{2}=1\\
y=\pm 1 & & y=\pm 1
\end{array}\right]$
The solution set is
$\{\ (2,1),\ (2, -1),\ (-2,1),\ (-2, -1)\ \}$
