Solve the systems of equations algebraically.
y = x^2 + 2x
y = 3x + 20
To solve this system of equations algebraically, we can use substitution. Since both equations are solved for y, we can set the right sides of the equations equal to each other:
x^2 + 2x = 3x + 20
Next, we want to get all the terms on one side of the equation to set it equal to 0:
x^2 - x - 20 = 0
This is a quadratic equation, so we'll try to factor it:
(x - 5)(x + 4) = 0
Now we can set each factor equal to 0 to find the possible values of x:
x - 5 = 0 or x + 4 = 0
x = 5 or x = -4
We have two potential solutions for x. Now we need to find the corresponding y-values. We can substitute each x-value back into either of the original equations (since they're both equal to y). We'll use the first equation:
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
So our two solutions are (5, 35) and (-4, 8).