Answer
The numbers are, $2\ \text{and 1}$, $2\ \text{ and }-1$, $-2\ \text{and 1}$ and $-2\ \text{ and }-\text{1}$
Work Step by Step
Let the numbers be $ x $ and $ y $ such that:
$\begin{align}
& {{x}^{2}}-{{y}^{2}}=3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{I} \right) \\
& 2{{x}^{2}}+{{y}^{2}}=9\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left( \text{II} \right)
\end{align}$
By adding equation (I) and equation (II): we get,
$\begin{align}
& {{x}^{2}}-{{y}^{2}}=3 \\
& \underline{2{{x}^{2}}+{{y}^{2}}=9} \\
& 3{{x}^{2}}=12 \\
& {{x}^{2}}=4 \\
\end{align}$
Now,
$ x=\pm 2$
Substitute $ x=2$ in equation (II):
$\begin{align}
& 2{{\left( 2 \right)}^{2}}+{{y}^{2}}=9 \\
& {{y}^{2}}=1 \\
& y=\pm 1
\end{align}$
Substitute $ x=-2$ in equation (II):
$\begin{align}
& 2{{\left( -2 \right)}^{2}}+{{y}^{2}}=9 \\
& {{y}^{2}}=1 \\
& y=\pm 1
\end{align}$
Thus, the numbers are, $2\ \text{and 1}$, $2\ \text{ and }-1$, $-2\ \text{and 1}$ or $-2\ \text{ and }-\text{1}$.
