note: evter 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 the system of equations algebraically, we can set both equations equal to each other:
x^2 + 2x = 3x + 20
Now, let's rearrange the equation to put it in standard quadratic form:
x^2 + 2x - 3x - 20 = 0
Combining like terms:
x^2 - x - 20 = 0
Now, we can factor the equation:
(x - 5)(x + 4) = 0
Setting each factor equal to zero:
x - 5 = 0 or x + 4 = 0
Solving for x:
x = 5 or x = -4
Now, we substitute these values of x back into either equation to find the corresponding values of y.
For x = 5:
y = (5)^2 + 2(5)
y = 25 + 10
y = 35
So, one solution is (x, y) = (5, 35).
For x = -4:
y = (-4)^2 + 2(-4)
y = 16 - 8
y = 8
So, the other solution is (x, y) = (-4, 8).
Therefore, the system of equations has two solutions: (5, 35) and (-4, 8).