Answer
$\left(-1,-\dfrac{1}{3}\right), \left(-1,\dfrac{1}{3}\right),\left(1,-\dfrac{1}{3}\right),\left(1,\dfrac{1}{3}\right)$
Work Step by Step
We are given the system:
$\begin{cases}
\dfrac{2}{x^2}+\dfrac{1}{y^2}=11\\
\dfrac{4}{x^2}-\dfrac{2}{y^2}=-14
\end{cases}$
We will use the addition method. Multiply Equation 1 by 2 and add it to Equation 2 to eliminate $y$ and determine $x$:
$\begin{cases}
2\left(\dfrac{2}{x^2}+\dfrac{1}{y^2}\right)=2(11)\\
\dfrac{4}{x^2}-\dfrac{2}{y^2}=-14
\end{cases}$
$2\left(\dfrac{2}{x^2}+\dfrac{1}{y^2}\right)+\dfrac{4}{x^2}-\dfrac{2}{y^2}=2(11)+(-14)$
$\dfrac{4}{x^2}+\dfrac{2}{y^2}+\dfrac{4}{x^2}-\dfrac{2}{y^2}=8$
$\dfrac{8}{x^2}=8$
$x^2=1$
$x_1=-1$
$x_2=1$
Substitute each of the values of $x$ in Equation 1 to determine $y$:
$\dfrac{2}{x^2}+\dfrac{1}{y^2}=11$
$x_1=-1\Rightarrow \dfrac{2}{(-1)^2}+\dfrac{1}{y^2}=11\Rightarrow \dfrac{1}{y^2}=9\Rightarrow y^2=\dfrac{1}{9}\Rightarrow y_1=-\dfrac{1}{3},y_2=\dfrac{1}{3}$
$x_2=1\Rightarrow \dfrac{2}{1^2}+\dfrac{1}{y^2}=11\Rightarrow \dfrac{1}{y^2}=9\Rightarrow y^2=\dfrac{1}{9}\Rightarrow y_3=-\dfrac{1}{3},y_4=\dfrac{1}{3}$
The system's solutions are:
$\left(-1,-\dfrac{1}{3}\right), \left(-1,\dfrac{1}{3}\right),\left(1,-\dfrac{1}{3}\right),\left(1,\dfrac{1}{3}\right)$
