## Precalculus (6th Edition) Blitzer

Solution is, $\left( 0,-2 \right)\text{ and }\left( 8,6 \right)$.
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.