We are given the following system of equations:
y = x^2 + 2x
y = 3x + 20
We can solve this system by substituting the first equation into the second equation for y:
x^2 + 2x = 3x + 20
Next, we can simplify by moving all the terms to one side of the equation:
x^2 - x - 20 = 0
We can then factor the quadratic equation to get:
(x - 5)(x + 4) = 0
Using the zero product property, we can solve for x:
x - 5 = 0 or x + 4 = 0
x = 5 or x = -4
Now that we have values for x, we can substitute them into either of the original equations to solve for y:
If x = 5, then y = 5^2 + 2(5) = 35
If x = -4, then y = (-4)^2 + 2(-4) = 8
Therefore, the solutions to the system are (5, 35) and (-4, 8).