Answer
$(6,1), (0,-2)$
Work Step by Step
We are given the system:
$\begin{cases}
x-2y=4\\
2y^2+xy=8
\end{cases}$
We will use the substitution method. Solve Equation 1 for $x$ and substitute the expression of $x$ in Equation 2 to eliminate $x$ and determine $y$:
$\begin{cases}
x=2y+4\\
2y^2+(2y+4)y=8
\end{cases}$
$2y^2+(2y+4)y=8$
$2y^2+2y^2+4y=8$
$4y^2+4y=8$
$4y^2+4y-8=0$
$4(y^2+y-2)=0$
$y^2+y-2=0$
$y^2-y+2y-2=0$
$y(y-1)+2(y-1)=0$
$(y-1)(y+2)=0$
$y-1=0\Rightarrow y_1=1$
$y+2=0\Rightarrow y_2=-2$
Substitute each of the values of $y$ in the expression of $x$ to determine $x$:
$x=2y+4$
$y_1=1\Rightarrow x_1=2(1)+4=6$
$y_2=-2\Rightarrow x_2=2(-2)+4=0$
The system's solutions are:
$(6,1), (0,-2)$
