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

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: \(2=1+\left(y-2\right)^{2}\) - step1: Swap the sides: \(1+\left(y-2\right)^{2}=2\) - step2: Add or subtract both sides: \(\left(y-2\right)^{2}=1\) - 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 \( y = 3 \).