## Precalculus (6th Edition) Blitzer

The second equation is multiplied by $-1$ and then added to the second equation to obtain ${{y}^{2}}+y=6$; it is a quadratic equation in $y$.
Let us consider the provided system: \begin{align} & {{x}^{2}}+{{y}^{2}}=13 \\ & {{x}^{2}}-y=7 \end{align} We multiply the second equation by $-1$ and add to the first equation: \begin{align} & {{x}^{2}}+{{y}^{2}}=13 \\ & \underline{-{{x}^{2}}+y=-7} \\ & {{y}^{2}}+y=6 \\ \end{align} Thus, the second equation multiplies by $-1$ and becomes ${{y}^{2}}+y=6$, a quadratic equation of $y$.