Answer
$(-2,-1), (-2,1), (2,-1),(2,1)$
Work Step by Step
We are given the system:
$\begin{cases}
2x^2-y^2=7\\
3x^2+2y^2=14
\end{cases}$
We will use the addition method. Multiply Equation 1 by 2 and add it to Equation 2 to eliminate $y$ and determine $x$:
$\begin{cases}
2(2x^2-y^2)=2(7)\\
3x^2+2y^2=14
\end{cases}$
$2(2x^2-y^2)+3x^2+2y^2=2(7)+14$
$4x^2-2y^2+3x^2+2y^2=28$
$7x^2=28$
$x^2=\dfrac{28}{7}$
$y=0\Rightarrow y_1=0$
$y-2=0\Rightarrow y_2=2$
$x^2=4$
$x=\pm\sqrt 4$
$x=\pm 2$
$x_1=-2$
$x_2=2$
Substitute each of the values of $x$ in Equation 1 to determine $y$:
$2x^2-y^2=7$
$x_1=-2\Rightarrow 2(-2)^2-y^2=7\Rightarrow y^2=1\Rightarrow y_1=-1,y_2=1$
$x_2=2\Rightarrow 2(2)^2-y^2=7\Rightarrow y^2=1\Rightarrow y_3=-1,y_4=1$
The system's solutions are:
$(-2,-1), (-2,1), (2,-1),(2,1)$
