Answer
The solution is, $\left\{ \left( 4,0 \right),\left( -3,\sqrt{7} \right)\text{ and }\left( -3,-\sqrt{7} \right) \right\}$
Work Step by Step
Let us consider the given equation:
$ x+{{y}^{2}}=4$ …… (I)
${{x}^{2}}+{{y}^{2}}=16$ …… (II)
Substract equation (I) from equation (II):
$\begin{align}
& {{x}^{2}}+{{y}^{2}}=16 \\
& x+{{y}^{2}}=4 \\
& \underline{-\text{ }-\text{ }-} \\
& {{x}^{2}}-x=12 \\
\end{align}$
Further simplify,
$\begin{align}
& {{x}^{2}}-x-12=0 \\
& x\left( x-4 \right)+3\left( x-4 \right)=0 \\
& \left( x-4 \right)\left( x+3 \right)=0 \\
& x=4,-3\,
\end{align}$
Substitute $ x=4$ in equation (I):
$\begin{align}
& 4+{{y}^{2}}=4 \\
& {{y}^{2}}=0 \\
& y=0
\end{align}$
Substitute $ x=-3$ in equation (II):
$\begin{align}
& -3+{{y}^{2}}=4 \\
& {{y}^{2}}=7 \\
& y=\pm \sqrt{7}
\end{align}$
Thus, the solution is $\left\{ \left( 4,0 \right),\left( -3,\sqrt{7} \right)\text{ and }\left( -3,-\sqrt{7} \right) \right\}$.
