13. $$ x=1+(y-2)^{2}, \quad x=2 $$ $$ 2-\left[1+(y-2)^{2}\right]=1-(y-2)^{2}=1-\left(y^{2}-4 y+4\right)=-y^{2}+4 y-3,50 $$

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To solve the given system of equations, we will first substitute the value of $$ x $$ from the first equation into the second equation.

Given equations:
1. $$ x = 1 + (y - 2)^2 $$
2. $$ x = 2 $$

Substitute $$ x = 2 $$ into the first equation:
$$ 2 = 1 + (y - 2)^2 $$

Now, we will solve this equation to find the value of $$ y $$.
Solve the quadratic equation by following steps:
- step0: Solve using square roots:
  $$2=1+\left(y-2ight)^{2}$$
- step1: Swap the sides:
  $$1+\left(y-2ight)^{2}=2$$
- step2: Add or subtract both sides:
  $$\left(y-2ight)^{2}=1$$
- step3: Simplify the expression:
  $$y-2=\pm \sqrt{1}$$
- step4: Simplify the expression:
  $$y-2=\pm 1$$
- step5: Separate into possible cases:
  $$\begin{align}&y-2=1\&y-2=-1\end{align}$$
- step6: Solve the equation:
  $$\begin{align}&y=3\&y=1\end{align}$$
- step7: Rewrite:
  $$y_{1}=1,y_{2}=3$$
The solutions to the equation are $$ y = 1 $$ and $$ y = 3 $$.
$$ y = 1 $$ or $$ y = 3 $$.

13. \( x=1+(y-2)^{2}, \quad x=2 \) \( 2-\left[1+(y-2)^{2}\right]=1-(y-2)^{2}=1-\left(y^{2}-4 y+4\right)=-y^{2}+4 y-3,50 \)

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