Solve the system of equations algebraically. Show all of your steps. y=x2+2x y=3x+20\
To solve the system of equations algebraically, we will set the two equations equal to each other and solve for x.
x^2 + 2x = 3x + 20
Next, we will subtract 3x and 20 from both sides to set the equation equal to zero:
x^2 - x - 20 = 0
Now, we will factor the quadratic equation:
(x - 5)(x + 4) = 0
Setting each factor equal to zero, we get:
x - 5 = 0 or x + 4 = 0
x = 5 or x = -4
Now that we have found the values of x, we can substitute them back into one of the original equations to solve for y:
Using x = 5:
y = (5)^2 + 2(5) = 25 + 10 = 35
So, one solution to the system of equations is x = 5, y = 35.
Using x = -4:
y = (-4)^2 + 2(-4) = 16 - 8 = 8
So, the other solution to the system of equations is x = -4, y = 8.
Therefore, the system of equations has two solutions: (5, 35) and (-4, 8).