Write this like a student and make it more concise: To solve this system of equations algebraically, we will set the two equations equal to each other and solve for x.
x^2 + 2x = 3x + 20
Next, we will bring all terms to one side to set the equation equal to zero:
x^2 + 2x - 3x - 20 = 0
x^2 - x - 20 = 0
Now we will factor the quadratic equation:
(x - 5)(x + 4) = 0
Setting each factor to zero:
x - 5 = 0 or x + 4 = 0
x = 5 or x = -4
Now that we have found the possible values of x, we will substitute them back into one of the original equations to find the corresponding values of y.
For x = 5:
y = (5)^2 + 2(5)
y = 25 + 10
y = 35
For x = -4:
y = (-4)^2 + 2(-4)
y = 16 - 8
y = 8
Therefore, the solutions to the system of equations are x = 5, y = 35 and x = -4, y = 8.