Solve the systems of equations algebraically.

y = x^2 + 2x
y = 3x + 20

1 answer

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).