Solve the following system by the substitution method. $$ \begin{array}{l}x+y=4 \ y=x^{2}-16\end{array} $$

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Solve the system of equations $$ x+y=4;y=x^{2}-16 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}x+y=4\y=x^{2}-16\end{array}ight.$$
- step1: Substitute the value of $$y:$$
  $$x+x^{2}-16=4$$
- step2: Move the expression to the left side:
  $$x+x^{2}-16-4=0$$
- step3: Subtract the numbers:
  $$x+x^{2}-20=0$$
- step4: Factor the expression:
  $$\left(x-4ight)\left(x+5ight)=0$$
- step5: Separate into possible cases:
  $$\begin{align}&x-4=0\&x+5=0\end{align}$$
- step6: Solve the equation:
  $$\begin{align}&x=4\&x=-5\end{align}$$
- step7: Calculate:
  $$x=4\cup x=-5$$
- step8: Rearrange the terms:
  $$\left\{ \begin{array}{l}x=4\y=x^{2}-16\end{array}ight.\cup \left\{ \begin{array}{l}x=-5\y=x^{2}-16\end{array}ight.$$
- step9: Calculate:
  $$\left\{ \begin{array}{l}x=4\y=0\end{array}ight.\cup \left\{ \begin{array}{l}x=-5\y=9\end{array}ight.$$
- step10: Calculate:
  $$\left\{ \begin{array}{l}x=-5\y=9\end{array}ight.\cup \left\{ \begin{array}{l}x=4\y=0\end{array}ight.$$
- step11: Check the solution:
  $$\left\{ \begin{array}{l}x=-5\y=9\end{array}ight.\cup \left\{ \begin{array}{l}x=4\y=0\end{array}ight.$$
- step12: Rewrite:
  $$\left(x,yight) = \left(-5,9ight)\cup \left(x,yight) = \left(4,0ight)$$
The solution to the system of equations by the substitution method is $$ (x,y) = (-5,9) $$ or $$ (x,y) = (4,0) $$.
The solutions are $$ (x,y) = (-5,9) $$ or $$ (x,y) = (4,0) $$.

Solve the following system by the substitution method.

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