## Precalculus (6th Edition) Blitzer

The solutions of the equation are $\left( -\frac{1}{2},\frac{1}{2} \right),\left( 2,8 \right)$
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 $\left( -\frac{1}{2},\frac{1}{2} \right),\left( 2,8 \right)$