Note: Enter your answer and show all the steps that you use to solve this problem in the space provided. Solve the system of equations algebraically. Show all of your steps.
y=x^2+2x
y=3x+20
To solve this system of equations, we will set the two equations equal to each other:
x^2 + 2x = 3x + 20
Next, we will rearrange the equation to set it 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 and solving for x, we get:
x - 5 = 0 or x + 4 = 0
x = 5 x = -4
Now that we have the x-values, we can plug them back into one of the original equations to find the corresponding y-values. Let's use y = x^2 + 2x:
If x = 5:
y = 5^2 + 2(5) = 25 + 10 = 35
So, one solution is (5, 35).
If x = -4:
y = (-4)^2 + 2(-4) = 16 - 8 = 8
So, the other solution is (-4, 8).
Therefore, the solutions to the system of equations are (5, 35) and (-4, 8).