Answer
$(-3,-2), (-3,2), (3,-2), (3,3)$
Work Step by Step
We are given the system:
$\begin{cases}
2x^2-5y^2=-2\\
3x^2+2y^2=35
\end{cases}$
We will use the addition method. Multiply Equation 1 by -3, Equation 2 by 2 and add them to eliminate $y$ and determine $x$:
$\begin{cases}
-3(2x^2-5y^2)=-3(-2)\\
2(3x^2+2y^2)=2(35)
\end{cases}$
$\begin{cases}
-6x^2+15y^2=6\\
6x^2+4y^2=70
\end{cases}$
$-6x^2+15y^2+6x^2+4y^2=6+70$
$19y^2=76$
$y^2=4$
$y=\pm 2$
$y_1=-2$
$y_2=2$
Substitute each of the values of $y$ in Equation 1 to determine $x$:
$2x^2-5y^2=-2$
$y_1=-2\Rightarrow 2x^2-5(-2)^2=-2\Rightarrow 2x^2=18\Rightarrow x^2=9\Rightarrow x_1=-3,x_2=3$
$y_2=2\Rightarrow 2x^2-5(2)^2=-2\Rightarrow 2x^2=18\Rightarrow x^2=9\Rightarrow x_3=-3,x_4=3$
The system's solutions are:
$(-3,-2), (-3,2), (3,-2), (3,3)$
