Solve the system of equations algebraically. Show all of your steps.
y=x2+2x
y=3x+20
To solve this system of equations algebraically, we will substitute the expression for y from the first equation into the second equation. This will allow us to solve for x.
Given equations:
1) y = x^2 + 2x
2) y = 3x + 20
Since both equations are equal to y, we can set them equal to each other:
x^2 + 2x = 3x + 20
Now, we need to rearrange this equation to solve for x:
x^2 + 2x - 3x - 20 = 0
x^2 - x - 20 = 0
Now, we need to factor this quadratic equation:
(x - 5)(x + 4) = 0
Setting each factor to zero to solve for x:
x - 5 = 0 or x + 4 = 0
x = 5 or x = -4
Now that we have found the possible values of x, we can substitute these values back into one of the original equations to find the corresponding values of y.
For x = 5:
y = (5)^2 + 2(5) = 25 + 10 = 35
So, (5, 35) is one solution.
For x = -4:
y = (-4)^2 + 2(-4) = 16 - 8 = 8
So, (-4, 8) is the other solution.
Therefore, the system of equations has two solutions: (5, 35) and (-4, 8).