Solve the system of equations. \[ \begin{array}{l}y=x^{2}-6 x-10 \\ y=-6 x-1\end{array} \] Write the coordinates in exact form. Simplify all (,\( \square \)

To solve the system of equations, we can substitute the second equation into the first. The equations are: 1. \(y = x^2 - 6x - 10\) 2. \(y = -6x - 1\) Setting the equations equal to each other gives us: \[ -6x - 1 = x^2 - 6x - 10 \] Now, we can simplify this equation: \[ 0 = x^2 - 6x + 6x - 10 + 1 \\ 0 = x^2 - 9 \] Next, we can factor the equation: \[ 0 = (x - 3)(x + 3) \] Thus, \(x = 3\) or \(x = -3\). Now, we can find the corresponding \(y\) values for each \(x\): 1. For \(x = 3\): \[ y = -6(3) - 1 = -18 - 1 = -19 \] So, one solution is \((3, -19)\). 2. For \(x = -3\): \[ y = -6(-3) - 1 = 18 - 1 = 17 \] So, the other solution is \((-3, 17)\). The final solutions are: \[ (3, -19) \quad \text{and} \quad (-3, 17) \]