Solving Quadratic Systems

We want to solve the given system of equations using the Substitution Method. $$\{\begin{array}{ll}y-\frac{1}{2}{x}^2=1+3x& \text{(I)}\\ y+\frac{1}{2}{x}^2=x& \text{(II)}\end{array}$$ Note that neither of the variables is isolated in either equation, so we will start by isolating $y$ in Equation (II). $$\begin{array}{c}\{\begin{array}{l}y-\frac{1}{2}{x}^2=1+3x\\ y+\frac{1}{2}{x}^2=x\end{array}\Rightarrow \{\begin{array}{l}y-\frac{1}{2}{x}^2=1+3x\\ y=\text{-}\frac{1}{2}{x}^2+x\end{array}\end{array}$$ The $y\text{-}$variable is isolated in Equation (II). This allows us to substitute its value $\text{-}\frac{1}{2}{x}^2+x$ for $y$ in Equation (I). Notice that in Equation (I), we have found that $x=\text{-}1.$ We can substitute this result into Equation (II) and solve for $y.$ We found that $y=\text{-}\frac{3}{2}.$ It means that the only solution to our system, which is the only point of intersection of the two parabolas, is $(\text{-}1,\text{-}\frac{3}{2}).$