To solve this system of equations algebraically, we will set the two equations equal to each other and solve for x.
So, we set x^2 + 2x = 3x + 20.
First, we'll move all the terms to one side of the equation to set it equal to zero:
x^2 + 2x - 3x - 20 = 0
x^2 - x - 20 = 0
Next, we'll factor the quadratic equation:
(x - 5)(x + 4) = 0
Setting each factor equal to zero gives us the possible values of x:
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. Let's choose y = x^2 + 2x:
For x = 5:
y = 5^2 + 2(5) = 25 + 10 = 35
So, when x = 5, y = 35.
For x = -4:
y = (-4)^2 + 2(-4) = 16 - 8 = 8
So, when x = -4, y = 8.
Therefore, the solution to the system of equations is:
x = 5, y = 35
x = -4, y = 8.
Solve the system of equations algebraically. Show all of your steps.
y=x^2+2x
y=3x+20
Can I get a good explanation? Like I know how the steps look after the first breakdown but by the second Nobody exactly explains what was added or subtracted to get to it.
1 answer