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 the system of equations algebraically, we need to find the values of x and y that satisfy both equations.
1. Start by setting the two expressions for y equal to each other, since they both represent y:
x^2 + 2x = 3x + 20
2. Rearrange the equation to bring all terms to one side:
x^2 + 2x - 3x - 20 = 0
3. Combine like terms:
x^2 - x - 20 = 0
4. Factor the quadratic equation (if possible) or use the quadratic formula:
(x - 5)(x + 4) = 0
This gives us two possible solutions for x:
x - 5 = 0 ---> x = 5
x + 4 = 0 ---> x = -4
5. Substitute each value of x back into one of the original equations to find the corresponding y-values:
For x = 5:
y = (5)^2 + 2(5) = 25 + 10 = 35
So one solution is (x, y) = (5, 35).
For x = -4:
y = (-4)^2 + 2(-4) = 16 - 8 = 8
So the other solution is (x, y) = (-4, 8).
Therefore, the system of equations has two solutions: (5, 35) and (-4, 8).
y=x^2+2x
y=3x+20
well that means that
x^2+2x = 3x+20
or
x^2 - x - 20 = 0
(x-5)(x+4) = 0
x = 5 or x = -4
if x = 5, y= 3x+20 = 35
if x = -4, y = 3x+20 = 8