Answer
$\{(\sqrt{5},\sqrt{2}),\ (\sqrt{5},-\sqrt{2}),\ (-\sqrt{5},\sqrt{2}),\ (-\sqrt{5},-\sqrt{2})\}$
Work Step by Step
1. Eliminating:
Eliminate the $(y^{2})$ terms:
$\left\{\begin{array}{lll}
16x^{2}-4y^{2}=72 & , & /\\
x^{2}-y^{2}=3 & , & /\times(-4), add
\end{array}\right.$
$(16-4)x^{2}=72-12$
$12x^{2}=60$
2. Solving:
$12x^{2}=60\qquad/\div 12$
$x^{2}=5$
$x=\pm\sqrt{5}$
3. Back-substituting into $x^{2}-y^{2}=3$:
$\left[\begin{array}{lll}
x=\sqrt{5} & or & x=-\sqrt{5}\\
5-y^{2}=3 & & 5-y^{2}=3 \\
-y^{2}=-2 & & -y^{2}=-2\\
y^{2}=2 & & y^{2}=2\\
y=\pm\sqrt{2} & & y=\pm\sqrt{2}
\end{array}\right]$
The solution set is
$\{(\sqrt{5},\sqrt{2}),\ (\sqrt{5},-\sqrt{2}),\ (-\sqrt{5},\sqrt{2}),\ (-\sqrt{5},-\sqrt{2})\}$
