Answer
Solution is, $\left( 0,-2 \right)\text{ and }\left( 8,6 \right)$.
Work Step by Step
Consider the provided system of equations,
$x-y=2$ …… (I)
${{y}^{2}}=4x+4$ …… (II)
Solve equation (I) for $y$ in terms of $x$ to get,
That is,
$\begin{align}
& x-y=2 \\
& y=x-2
\end{align}$
Now by substituting the value of $y$ obtained above in equation (II), we get,
Thus,
$\begin{align}
& {{\left( x-2 \right)}^{2}}=4x+4 \\
& {{x}^{2}}-4x+4=4x+4
\end{align}$
Add $-4x-4$ on both sides,
$\begin{align}
& {{x}^{2}}-4x+4-4x-4=4x+2-4x-4 \\
& {{x}^{2}}-8x=0 \\
& x\left( x-8 \right)=0
\end{align}$
This implies:
Either $x=0$ or, $x-8=0$, that is, $x=8$
Now, from equation (I),
If $x=0$,
Then,
$\begin{align}
& 0-y=2 \\
& y=-2
\end{align}$
If $x=8$ ,
Then,
$\begin{align}
& 8-y=2 \\
& y=8-2 \\
& y=6
\end{align}$
Hence, $\left( 0,-2 \right)$ and $\left( 8,6 \right)$ are the solutions of the provided system.
