Solve the system of equations. \[ y=x^{2}-2 x-20 \] \( y=-6 x+1 \) Write the coordinates in exact form. Simplify all fractions and rad (

To solve the system of equations, we will set the two equations equal to each other since they both equal \(y\). Starting with: \[ x^2 - 2x - 20 = -6x + 1 \] Rearranging this gives: \[ x^2 - 2x + 6x - 20 - 1 = 0 \] This simplifies to: \[ x^2 + 4x - 21 = 0 \] Next, we can factor this equation: \[ (x + 7)(x - 3) = 0 \] Setting each factor to zero gives us: \[ x + 7 = 0 \quad \Rightarrow \quad x = -7 \] \[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \] Now we substitute \(x = -7\) and \(x = 3\) back into one of the original equations to find the corresponding \(y\) values. For \(x = -7\): \[ y = -6(-7) + 1 = 42 + 1 = 43 \] So, one solution is \((-7, 43)\). For \(x = 3\): \[ y = -6(3) + 1 = -18 + 1 = -17 \] So, another solution is \((3, -17)\). The coordinates of the points of intersection in exact form are: \[ (-7, 43) \quad \text{and} \quad (3, -17) \] Now you have two beautiful points where these equations dance together on the coordinate plane!