Solve the system of equations. \[ \begin{array}{l}y=x^{2}-24 x+22 \\ y=-32 x+10 \\ \text { Write the coordinates in exact form. Simplify al } \\ \text { ( , }\end{array} \]

Solve the system of equations \( y=x^{2}-24x+22;y=-32x+10 \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}y=x^{2}-24x+22\\y=-32x+10\end{array}\right.\) - step1: Substitute the value of \(y:\) \(x^{2}-24x+22=-32x+10\) - step2: Move the expression to the left side: \(x^{2}-24x+22-\left(-32x+10\right)=0\) - step3: Calculate: \(x^{2}+8x+12=0\) - step4: Factor the expression: \(\left(x+2\right)\left(x+6\right)=0\) - step5: Separate into possible cases: \(\begin{align}&x+2=0\\&x+6=0\end{align}\) - step6: Solve the equation: \(\begin{align}&x=-2\\&x=-6\end{align}\) - step7: Calculate: \(x=-2\cup x=-6\) - step8: Rearrange the terms: \(\left\{ \begin{array}{l}x=-2\\y=x^{2}-24x+22\end{array}\right.\cup \left\{ \begin{array}{l}x=-6\\y=x^{2}-24x+22\end{array}\right.\) - step9: Calculate: \(\left\{ \begin{array}{l}x=-2\\y=74\end{array}\right.\cup \left\{ \begin{array}{l}x=-6\\y=202\end{array}\right.\) - step10: Check the solution: \(\left\{ \begin{array}{l}x=-2\\y=74\end{array}\right.\cup \left\{ \begin{array}{l}x=-6\\y=202\end{array}\right.\) - step11: Rewrite: \(\left(x,y\right) = \left(-2,74\right)\cup \left(x,y\right) = \left(-6,202\right)\) The solutions to the system of equations are \( (x, y) = (-2, 74) \) and \( (x, y) = (-6, 202) \). Therefore, the coordinates in exact form are \( (-2, 74) \) and \( (-6, 202) \).