Solve the system of equations. \[ y=28 x+18 \] \( y=x^{2}+45 x+48 \) Write the coordinates in exact form. Simpli \( (\square \)

To solve the system of equations, we set the two equations equal to each other: \[ 28x + 18 = x^2 + 45x + 48. \] Rearranging gives us: \[ x^2 + 45x + 48 - 28x - 18 = 0, \] which simplifies to: \[ x^2 + 17x + 30 = 0. \] Now, we can factor this quadratic equation: \[ (x + 2)(x + 15) = 0. \] Setting each factor to zero gives us the solutions for \(x\): \[ x + 2 = 0 \quad \Rightarrow \quad x = -2, \] \[ x + 15 = 0 \quad \Rightarrow \quad x = -15. \] Now, we substitute these \(x\) values back into either equation to find corresponding \(y\) values. Using the first equation, \(y = 28x + 18\): 1. For \(x = -2\): \[ y = 28(-2) + 18 = -56 + 18 = -38. \] Thus, one point is \((-2, -38)\). 2. For \(x = -15\): \[ y = 28(-15) + 18 = -420 + 18 = -402. \] Thus, the other point is \((-15, -402)\). Putting it all together, the exact coordinates are: \[ (-2, -38) \quad \text{and} \quad (-15, -402). \]