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

To solve the system of equations, we need to set the two equations equal to each other since both equal \(y\): \[ x^2 - 24x + 22 = -32x + 10 \] Now, let's rearrange the equation: \[ x^2 - 24x + 32x + 22 - 10 = 0 \] This simplifies to: \[ x^2 + 8x + 12 = 0 \] Next, we will factor the quadratic equation: \[ (x + 6)(x + 2) = 0 \] Setting each factor to zero gives us the solutions for \(x\): \[ x + 6 = 0 \quad \Rightarrow \quad x = -6 \] \[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \] Now, we will find the corresponding \(y\) values using the second equation \(y = -32x + 10\): 1. For \(x = -6\): \[ y = -32(-6) + 10 = 192 + 10 = 202 \] 2. For \(x = -2\): \[ y = -32(-2) + 10 = 64 + 10 = 74 \] Thus, the coordinates of the solutions are: \[ (-6, 202) \quad \text{and} \quad (-2, 74) \] Therefore, the exact solutions are: \[ (-6, 202) \quad \text{and} \quad (-2, 74) \]