Answer
$(-2,1), (2,-1)$
Work Step by Step
We are given the system:
$\begin{cases}
2x^2+xy=6\\
x^2+2xy=0
\end{cases}$
We will use the addition method. Multiply Equation 1 by -2 and add it to Equation 2 to eliminate $y$ and determine $x$:
$\begin{cases}
-2(2x^2+xy)=-2(6)\\
x^2+2xy=0
\end{cases}$
$-2(2x^2+xy)+x^2+2xy=-12+0$
$-4x^2-2xy+x^2+2xy=-12$
$-3x^2=-12$
$x^2=\dfrac{-12}{-3}$
$x^2=4$
$x=\pm\sqrt 4$
$x=\pm 2$
$x_1=-2$
$x_2=2$
Substitute each of the values of $x$ in Equation 2 to determine $y$:
$x^2+2xy=0$
$x_1=-2\Rightarrow (-2)^2+2(-2)y=0\Rightarrow 4-4y=0\Rightarrow 4=4y\Rightarrow y_1=1$
$x_2=2\Rightarrow 2^2+2(2)y=0\Rightarrow 4+4y=0\Rightarrow -4=4y\Rightarrow y_2=-1$
The system's solutions are:
$(-2,1), (2,-1)$
