Answer
$\left(-1,-\dfrac{1}{2}\right), \left(-1,\dfrac{1}{2}\right),\left(1,-\dfrac{1}{2}\right),\left(1,\dfrac{1}{2}\right)$
Work Step by Step
We are given the system:
$\begin{cases}
\dfrac{3}{x^2}+\dfrac{1}{y^2}=7\\
\dfrac{5}{x^2}-\dfrac{2}{y^2}=-3
\end{cases}$
We will use the addition method. Multiply Equation 1 by 2 and add it to Equation 2 to eliminate $y$ and determine $x$:
$\begin{cases}
2\left(\dfrac{3}{x^2}+\dfrac{1}{y^2}\right)=2(7)\\
\dfrac{5}{x^2}-\dfrac{2}{y^2}=-3
\end{cases}$
$2\left(\dfrac{3}{x^2}+\dfrac{1}{y^2}\right)+\dfrac{5}{x^2}-\dfrac{2}{y^2}=2(7)+(-3)$
$\dfrac{6}{x^2}+\dfrac{2}{y^2}+\dfrac{5}{x^2}-\dfrac{2}{y^2}=11$
$\dfrac{11}{x^2}=11$
$x^2=1$
$x_1=-1$
$x_2=1$
Substitute each of the values of $x$ in Equation 1 to determine $y$:
$\dfrac{3}{x^2}+\dfrac{1}{y^2}=7$
$x_1=-1\Rightarrow \dfrac{3}{(-1)^2}+\dfrac{1}{y^2}=7\Rightarrow \dfrac{1}{y^2}=4\Rightarrow y^2=\dfrac{1}{4}\Rightarrow y_1=-\dfrac{1}{2},y_2=\dfrac{1}{2}$
$x_2=1\Rightarrow \dfrac{3}{1^2}+\dfrac{1}{y^2}=7\Rightarrow \dfrac{1}{y^2}=4\Rightarrow y^2=\dfrac{1}{4}\Rightarrow y_3=-\dfrac{1}{2},y_4=\dfrac{1}{2}$
The system's solutions are:
$\left(-1,-\dfrac{1}{2}\right), \left(-1,\dfrac{1}{2}\right),\left(1,-\dfrac{1}{2}\right),\left(1,\dfrac{1}{2}\right)$
