Solve the system of equations algebraically. Show all of your steps.
y=x^2+2x
y=3x+20
Since both equations are equal to y, we can set them equal to each other and solve for x:
x^2 + 2x = 3x + 20
x^2 + 2x - 3x - 20 = 0
x^2 - x - 20 = 0
Next, factor the quadratic equation:
(x - 5)(x + 4) = 0
Setting each factor to zero and solving for x:
x - 5 = 0
x = 5
x + 4 = 0
x = -4
So, the solutions for x are x = 5 and x = -4.
Now, plug these values of x into either of the original equations to find the corresponding values of y:
For x = 5:
y = 3(5) + 20
y = 15 + 20
y = 35
So, when x = 5, y = 35.
For x = -4:
y = 3(-4) + 20
y = -12 + 20
y = 8
So, when x = -4, y = 8.
Therefore, the solutions to the system of equations are (5, 35) and (-4, 8).