Answer
The solution is $\left\{ \left( -1,-\frac{1}{3} \right),\left( -1,\frac{1}{3} \right),\left( 1,\frac{-1}{3} \right),\left( 1,\frac{1}{3} \right) \right\}$
Work Step by Step
The provided equations are,
$\begin{align}
& \frac{2}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}=11\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\
& \frac{4}{{{x}^{2}}}-\frac{2}{{{y}^{2}}}=-14\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right)
\end{align}$
Multiply the equation (I) by (II) and add the equation (I) and equation (II):
$\begin{align}
& \frac{4}{{{x}^{2}}}+\frac{2}{{{y}^{2}}}=22 \\
& \frac{4}{{{x}^{2}}}-\frac{2}{{{y}^{2}}}=-14 \\
& \overline{\begin{align}
& \,\,\,\,\,\,\frac{8}{{{x}^{2}}}=8\,\,\,\,\,\,\,\,\,\,\,\,\, \\
& x=\pm 1
\end{align}} \\
\end{align}$
Substitute the value of $ x=1,x=-1$ in equation (I)
For
$\begin{align}
& x=-1; \\
& \frac{2}{{{\left( -1 \right)}^{2}}}+\frac{1}{{{y}^{2}}}=11 \\
& 2+\frac{1}{{{y}^{2}}}=11 \\
& {{y}^{2}}=\frac{1}{9}
\end{align}$
Then,
$ y=\pm \frac{1}{3}$
For
$\begin{align}
& x=1; \\
& \frac{2}{{{\left( 1 \right)}^{2}}}+\frac{1}{{{y}^{2}}}=11 \\
& 2+\frac{1}{{{y}^{2}}}=11 \\
& {{y}^{2}}=\frac{1}{9}
\end{align}$
Then,
$ y=\pm \frac{1}{3}$
Thus, the solution set of system is $\left\{ \left( -1,-\frac{1}{3} \right),\left( -1,\frac{1}{3} \right),\left( 1,\frac{-1}{3} \right),\left( 1,\frac{1}{3} \right) \right\}$.
