Precalculus (6th Edition) Blitzer

The numbers are, $2\ \text{and 1}$, $2\ \text{ and }-1$, $-2\ \text{and 1}$ and $-2\ \text{ and }-\text{1}$
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 $-2\ \text{ and }-\text{1}$.