Answer
The solutions of the equation are $\left( -\frac{1}{2},\frac{1}{2} \right),\left( 2,8 \right)$
Work Step by Step
Let us consider the system of equations as $\begin{align}
& 3x-y=-2 \\
& 2{{x}^{2}}-y=0
\end{align}$.
Also, consider the equation $2{{x}^{2}}-y=0$:
$\begin{align}
& 2{{x}^{2}}-y=0 \\
& y=2{{x}^{2}}
\end{align}$
Now, put $ y=2{{x}^{2}}$ in $3x-y=-2$ and compute the values as given below:
$\begin{align}
& 3x-y=-2 \\
& 3x-2{{x}^{2}}=-2 \\
& 2{{x}^{2}}-3x-2=0 \\
& \left( 2x+1 \right)\left( x-2 \right)=0
\end{align}$
Solve further the equation:
$\begin{align}
& \left( 2x+1 \right)\left( x-2 \right)=0 \\
& x=-\frac{1}{2},2
\end{align}$
Substitute the value $ x=-\frac{1}{2}$ in the equation $ y=2{{x}^{2}}$, to obtain the value of y:
$\begin{align}
& y=2{{x}^{2}} \\
& y=2\times {{\left( -\frac{1}{2} \right)}^{2}} \\
& y=\frac{1}{2} \\
\end{align}$
Substitute the value $ x=2$ in the equation $ y=2{{x}^{2}}$, to obtain the value of y :
$\begin{align}
& y=2{{x}^{2}} \\
& y=2\times {{2}^{2}} \\
& y=8 \\
\end{align}$
Thus, the solution of equations are $\left( -\frac{1}{2},\frac{1}{2} \right),\left( 2,8 \right)$
