Answer
The solution is $\left\{ \left( -1,-\frac{1}{2} \right),\left( -1,\frac{1}{2} \right),\left( 1,\frac{-1}{2} \right),\left( 1,\frac{1}{2} \right) \right\}$
Work Step by Step
We have the equations,
$\begin{align}
& \frac{3}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}=7\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\
& \frac{5}{{{x}^{2}}}-\frac{2}{{{y}^{2}}}=-3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right)
\end{align}$
And multiply the equation (I) by 2 and then add the equation (I) and equation (II):
$\begin{align}
& \frac{6}{{{x}^{2}}}+\frac{2}{{{y}^{2}}}=14 \\
& \frac{5}{{{x}^{2}}}-\frac{2}{{{y}^{2}}}=-3 \\
& \overline{\begin{align}
& \,\,\,\,\,\,\frac{11}{{{x}^{2}}}=11\,\,\,\,\,\,\,\,\,\,\,\,\, \\
& {{x}^{2}}=1 \\
\end{align}} \\
\end{align}$
Finally,
$ x=\pm 1$
Substitute the value of x as $ x=1,x=-1$ into equation (I):
For
$\begin{align}
& x=-1; \\
& \frac{3}{{{\left( -1 \right)}^{2}}}+\frac{1}{{{y}^{2}}}=7 \\
& 3+\frac{1}{{{y}^{2}}}=7 \\
& {{y}^{2}}=\frac{1}{4}
\end{align}$
Then,
$ y=\pm \frac{1}{2}$
For
$\begin{align}
& x=1; \\
& \frac{3}{{{\left( 1 \right)}^{2}}}+\frac{1}{{{y}^{2}}}=7 \\
& 3+\frac{1}{{{y}^{2}}}=7 \\
& {{y}^{2}}=\frac{1}{4}
\end{align}$
Then,
$ y=\pm \frac{1}{2}$
Therefore, the solution set of system is $\left\{ \left( -1,-\frac{1}{2} \right),\left( -1,\frac{1}{2} \right),\left( 1,\frac{-1}{2} \right),\left( 1,\frac{1}{2} \right) \right\}$.
