Answer
$\left( 1,-1 \right),\left( 1,1 \right),\left( -1,1 \right),\left( -1,-1 \right)$
Work Step by Step
Consider the given expression,
${{x}^{2}}-2{{y}^{2}}=-1$ …… (1)
$2{{x}^{2}}-{{y}^{2}}=1$ …… (2)
Now, we will multiply with $2$ in equation (1),
$\begin{align}
& 2{{x}^{2}}-4{{y}^{2}}=-2 \\
& 2{{x}^{2}}-{{y}^{2}}=1
\end{align}$ …… (3)
Using (2) and (3), we get,
$\begin{align}
& 2{{x}^{2}}-4{{y}^{2}}=-2 \\
& 2{{x}^{2}}-{{y}^{2}}=1 \\
& -3{{y}^{2}}=-3
\end{align}$
Therefore,
${{y}^{2}}=\pm 1$
Now, substitute $y=1$ in equation (1). Then,
$\begin{align}
& {{x}^{2}}-2{{y}^{2}}=-1 \\
& {{x}^{2}}-2{{\left( 1 \right)}^{2}}=-1 \\
& x=\pm 1
\end{align}$
So, the solution of the provided expression is,
$\begin{align}
& x=\pm 1 \\
& y=\pm 1 \\
\end{align}$
