Answer
$x=4$ and $y=0$
Work Step by Step
$\begin{cases}
y+x^2=4x\\
y+4x=16
\end{cases}$
Multiplying the second system of equations by -1 and adding it to the first system of equations...
$\begin{cases}
y+x^2-4x=0\\
-y-4x=-16\\
-- -- -- -\\
x^2-8x=-16
\end{cases}$
Therefore, $x^2-8x+16=0$.
Solving for the trinomial using factoring, find two numbers whose sum is $-8$ whose multiple is $+16$. The numbers are $-4$ and $-4$.
$x^2-8x+16=x^2-4x-4x+16$,
$x(x-4)-4(x-4)=(x-4)(x-4)=(x-4)^2$.
Thus, $x=4$ and subsituting back in, $y=0$.
