Answer
$\left\{\left( 1,-2\right), \left( 4,4\right) \right\}$
Work Step by Step
Substituting $
y=x+2
$ into the 2nd equation results to
\begin{array}{l}
(2x-4)^2=4x\\\\
4x^2-16x+16=4x\\\\
4x^2-20x+16=0\\\\
x^2-5x+4=0\text{... divide both sides by $4$}\\\\
(x-4)(x-1)=0\\\\
x=\left\{
1,4
\right\}
.\end{array}
Substituting $x=1$ into the 1st equation results to
\begin{array}{l}
y=2(1)-4\\\\
y=-2
.\end{array}
Substituting $x=4$ into the 1st equation results to
\begin{array}{l}
y=2(4)-4\\\\
y=4
.\end{array}
Hence, the solution set is $
\left\{\left( 1,-2\right), \left( 4,4\right) \right\}
$
