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