Answer
The solution set is $\left\{ \left( 2,\sqrt{3} \right),\left( 2,-\sqrt{3} \right),\left( -2,\sqrt{3} \right),\left( -2.-\sqrt{3} \right) \right\}$.
Work Step by Step
Let us consider the provided system:
$\begin{align}
& {{x}^{2}}+4{{y}^{2}}=16 \\
& {{x}^{2}}-{{y}^{2}}=1
\end{align}$
And subtract the second equation form the first to obtain,
$\begin{align}
& {{x}^{2}}+4{{y}^{2}}=16 \\
& {{x}^{2}}-{{y}^{2}}=1 \\
& \underline{-+-} \\
\end{align}$
$\begin{align}
& 5{{y}^{2}}=15 \\
& {{y}^{2}}=3 \\
& y=\pm \sqrt{3}
\end{align}$
Substitute $y=\sqrt{3}$ in the second equation and solve for x:
$\begin{align}
& {{x}^{2}}-{{\left( \sqrt{3} \right)}^{2}}=1 \\
& {{x}^{2}}=4 \\
& x=\pm 2
\end{align}$
Substitute $y=-\sqrt{3}$ in the second equation and solve for x:
$\begin{align}
& {{x}^{2}}+{{\left( -\sqrt{3} \right)}^{2}}=1 \\
& {{x}^{2}}=4 \\
& x=\pm 2
\end{align}$
Thus, the solution set is $\left\{ \left( 2,\sqrt{3} \right),\left( 2,-\sqrt{3} \right),\left( -2,\sqrt{3} \right),\left( -2.-\sqrt{3} \right) \right\}$.
