for $$ x $$ and $$ y $$ $$ y=x+2 $$ $$ x+x y=4 $$

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Solve the system of equations $$ x+x*y=4;y=x+2 $$.
Solve the system of equations by following steps:
- step0: Solve using the substitution method:
  $$\left\{ \begin{array}{l}x+xy=4\y=x+2\end{array}ight.$$
- step1: Substitute the value of $$y:$$
  $$x+x\left(x+2ight)=4$$
- step2: Simplify:
  $$3x+x^{2}=4$$
- step3: Move the expression to the left side:
  $$3x+x^{2}-4=0$$
- step4: Factor the expression:
  $$\left(x-1ight)\left(x+4ight)=0$$
- step5: Separate into possible cases:
  $$\begin{align}&x-1=0\&x+4=0\end{align}$$
- step6: Solve the equation:
  $$\begin{align}&x=1\&x=-4\end{align}$$
- step7: Calculate:
  $$x=1\cup x=-4$$
- step8: Rearrange the terms:
  $$\left\{ \begin{array}{l}x=1\y=x+2\end{array}ight.\cup \left\{ \begin{array}{l}x=-4\y=x+2\end{array}ight.$$
- step9: Calculate:
  $$\left\{ \begin{array}{l}x=1\y=3\end{array}ight.\cup \left\{ \begin{array}{l}x=-4\y=-2\end{array}ight.$$
- step10: Calculate:
  $$\left\{ \begin{array}{l}x=-4\y=-2\end{array}ight.\cup \left\{ \begin{array}{l}x=1\y=3\end{array}ight.$$
- step11: Check the solution:
  $$\left\{ \begin{array}{l}x=-4\y=-2\end{array}ight.\cup \left\{ \begin{array}{l}x=1\y=3\end{array}ight.$$
- step12: Rewrite:
  $$\left(x,yight) = \left(-4,-2ight)\cup \left(x,yight) = \left(1,3ight)$$
The solutions to the system of equations are $$ (x,y) = (-4,-2) $$ and $$ (x,y) = (1,3) $$.
The solutions are $$ x = -4, y = -2 $$ and $$ x = 1, y = 3 $$.

for \( x \) and \( y \) \( y=x+2 \) \( x+x y=4 \)

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