Answer
-2 or 2 and -1 or 1
Work Step by Step
We can write the system:
$\begin{cases}
x^2-y^2=3\\
2x^2+y^2=9
\end{cases}$
We will use the addition method. Add Equation 2 to Equation 1 to eliminate $y$ and determine $x$:
$x^2-y^2+2x^2+y^2=3+9$
$3x^2=12$
$x^2=\dfrac{12}{3}$
$x^2=4$
$x=\sqrt 4$
$x=\pm 2$
$x_1=-2$
$x_2=2$
Substitute each of the values of $x$ in Equation 1 to determine $y$:
$x^2-y^2=3$
$x_1=-2\Rightarrow (-2)^2-y^2=3\Rightarrow y^2=1\Rightarrow y_1=-1,y_2=1$
$x_2=2\Rightarrow 2^2-y^2=3\Rightarrow y^2=1\Rightarrow y_1=-1,y_2=1$
The system's solutions are:
$(-2,-1), (-2,1), (2,-1),(2,1)$
The numbers are -2 or 2 and -1 or 1