Question

Enter your answer and show all the steps that you use to solve this problem

Solve the system of equations algerbaically show all your steps

Y=x^2+2x
Y=3x+20

Answer 1

To solve this system of equations algebraically, we set the two equations equal to each other:

x^2 + 2x = 3x + 20

Next, we bring all terms to one side of the equation:

x^2 + 2x - 3x - 20 = 0
x^2 - x - 20 = 0

Now, we need to factor the quadratic equation. We need two integers that multiply to -20 and add up to -1. The factors that satisfy this condition are -5 and 4:

(x - 5)(x + 4) = 0

Now we find the roots by setting each factor to zero:

x - 5 = 0 or x + 4 = 0
x = 5 or x = -4

Therefore, the solutions for x are x = 5 and x = -4.

Now, we substitute these values back into one of the original equations to find the corresponding y values. Let's use Y = x^2 + 2x:

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.