Answer
$(2,8),\left(-\dfrac{1}{2},\dfrac{1}{2}\right)$
Work Step by Step
We are given the system:
$\begin{cases}
3x-y=-2\\
2x^2-y=0
\end{cases}$
We will use the substitution method. Solve Equation 1 for $y$ and substitute the expression of $y$ in Equation 2 to eliminate $y$ and determine $x$:
$\begin{cases}
y=3x+2\\
2x^2-(3x+2)=0
\end{cases}$
$2x^2-3x-2=0$
$2x^2-4x+x-2=0$
$2x(x-2)+(x-2)=0$
$(x-2)(2x+1)=0$
$x-2=0\Rightarrow x_1=2$
$2x+1=0=0\Rightarrow x_2=-\dfrac{1}{2}$
Substitute each of the values of $x$ in the expression of $y$ to determine $y$:
$y=3x+2$
$x_1=2\Rightarrow y_1=3(2)+2=8$
$x_2=-\dfrac{1}{2}\Rightarrow y_2=3\left(-\dfrac{1}{2}\right)+2=\dfrac{1}{2}$
The system's solutions are:
$(2,8),\left(-\dfrac{1}{2},\dfrac{1}{2}\right)$
