Question

Solve the system algebraically, show steps.

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

Answer 1

To solve this system algebraically, we need to set the two equations equal to each other and then solve for x.

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

Now, we have a quadratic equation. We can solve this equation by factoring, completing the square, or using the quadratic formula. In this case, let's factor the equation.

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

Setting each factor equal to zero gives us the possible values for x:

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

Now that we have the values of x, we can substitute them back into one of the original equations to find the corresponding values of y. Let's use y = x^2 + 2x.

When x = 5:
y = (5)^2 + 2(5) = 25 + 10 = 35
Therefore, when x = 5, y = 35.

When x = -4:
y = (-4)^2 + 2(-4) = 16 - 8 = 8
Therefore, when x = -4, y = 8.

So, the solution to the system is x = 5, y = 35 and x = -4, y = 8.