how do you solve the system of equations: 2x-y=4 and y=x^2-6x+8

Question

2x-y=4

y=x^2-6x+8

Answer 1

Given equation 2x-y = 4

Subtrac 2x from each side.

2x-2x-y = 4-2x

-y  = 4-2x

Multiple to each side by negitive one.

y = 2x-4 ---------> (1)

And y = x^2-6x+8 -------> (2)

From (1) substitute the y value in (2).

2x-4 = x^2-6x+8

Bring all terms to one side.

2x-4-x^2+6x-8 = 0

-x^2+8x-12 = 0

x^2-8x+12 = 0

x^2-6x-2x+12 = 0

x(x-6)-2(x-6) = 0

(x-6)(x-2) = 0

x-6 = 0 and x-2 = 0

x = 6 and x = 2

Substitute the x values in y = 2x-4

For x = 6 ⇒ y = 2*6-4

y = 12-4

y  = 8

For x = 2 ⇒ y = 2*2-4

y = 4-4

y = 0

Solution x = 6, y = 8 and x = 2, y = 0.