Which of the following uses the substitution method to solve linear-quadratic system y=x2−4x+8, y=3x−4?

Answer 1

To solve this system using the substitution method, we will substitute the second equation y=3x-4 into the first equation y=x^2-4x+8.

So, we have:

x^2-4x+8 = 3x-4

Rearranging this equation, we get:

x^2 - 7x + 12 = 0

Now, we can factor this quadratic equation:

(x-3)(x-4) = 0

So, x = 3 or x = 4

Now, substituting these x-values back into the second equation y=3x-4, we get:

When x = 3, y = 3(3) - 4 = 5.

When x = 4, y = 3(4) - 4 = 8.

Therefore, the solutions to the system are x = 3, y = 5 and x = 4, y = 8.