Solve the system of equations algebraically. Show all of your steps.
y = x^2 + 2x
y = 3x + 20
To solve this system of equations algebraically, we will set these two equations equal to each other since they both equal y.
So we have:
x^2 + 2x = 3x + 20
Rearranging the terms, we get:
x^2 + 2x - 3x - 20 = 0
x^2 - x - 20 = 0
Now we need to factor the quadratic equation or solve it using the quadratic formula. In this case, the equation factors nicely:
(x - 5)(x + 4) = 0
Setting each factor to zero gives us two possible solutions for x:
x - 5 = 0 or x + 4 = 0
x = 5 or x = -4
Now that we have found the values of x, we can substitute them back into one of the original equations to find the corresponding y values. Let's use the first equation:
y = x^2 + 2x
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
x = -4, y = 8