Answer
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 $y$.
Work Step by Step
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 $y$.
