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Solve the system of equations algebraically. Show all of your steps.
y=x2+2x
y=3x+20
Since both equations are set equal to y, we can set them equal to each other:
x^2 + 2x = 3x + 20
Subtracting 3x and 20 from both sides, we get:
x^2 - x - 20 = 0
Factoring, we get:
(x - 5)(x + 4) = 0
Therefore, x = 5 or x = -4.
Substituting these values of x back into either equation, we can solve for y:
If x = 5:
y = 5^2 + 2(5) = 35
If x = -4:
y = (-4)^2 + 2(-4) = 8
Therefore, the solution to the system of equations is (5, 35) and (-4, 8).
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To solve the system of equations algebraically, we can set the two equations equal to each other since they both equal to y.
x^2 + 2x = 3x + 20
First, let's rearrange the equation by moving all the terms to one side:
x^2 + 2x - 3x - 20 = 0
Combine like terms:
x^2 - x - 20 = 0
Next, we need to solve the quadratic equation. One way to do this is by factoring, if possible.
The equation x^2 - x - 20 = 0 can be factored as (x - 5)(x + 4) = 0.
Setting each factor equal to 0, we get:
x - 5 = 0 or x + 4 = 0
Solving each equation for x, we have:
x = 5 or x = -4
Now that we have the values of x, we can substitute them back into either equation to find the corresponding values of y.
Using the first equation y = x^2 + 2x:
When x = 5, y = 5^2 + 2(5) = 25 + 10 = 35
When x = -4, y = (-4)^2 + 2(-4) = 16 - 8 = 8
Therefore, the solution to the system of equations is (x, y) = (5, 35) and (-4, 8).