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