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Solve the system of equations algebraically. Show all of your steps.
y = x²+2x
y = 3x + 20
To solve this system of equations algebraically, we will set the two equations equal to each other since they both equal y:
x² + 2x = 3x + 20
Next, we will move all terms to one side to set the equation equal to zero:
x² + 2x - 3x - 20 = 0
Simplify the equation:
x² - 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, plug these x-values back into either of the original equations to solve for 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
x = -4, y = 8