I am stuck on this.
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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 need to find the values of x and y that satisfy both equations simultaneously. The two equations are:
Equation 1: y = x^2 + 2x
Equation 2: y = 3x + 20
To find the solution, we can set the two equations equal to each other and solve for x:
x^2 + 2x = 3x + 20
Now, we have a quadratic equation. We can rearrange it to have all terms on one side:
x^2 + 2x - 3x - 20 = 0
Combining like terms:
x^2 - x - 20 = 0
Next, we can try to factor the quadratic equation or use the quadratic formula to find the values of x. Let's try factoring:
(x - 5)(x + 4) = 0
Now, we can set each factor equal to zero and solve for x:
x - 5 = 0
x = 5
x + 4 = 0
x = -4
So, we have two possible solutions for x: x = 5 and x = -4.
To find the corresponding y-values, we substitute each x-value back into one of the original equations (either Equation 1 or Equation 2). Let's use Equation 1:
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 and x = -4, y = 8.