Which of the following uses the substitution method to solve the linear-quadratic system y = x^2 - 4x + 8, y = 3x - 4?
The linear-quadratic system y = x^2 - 4x + 8, y = 3x - 4 can be solved using the substitution method by substituting the value of y from the second equation into the first equation.
Let's substitute y = 3x - 4 into y = x^2 - 4x + 8:
3x - 4 = x^2 - 4x + 8
Rearranging the terms, we get:
x^2 - 3x - 12 = 0
Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. Let's use factoring:
(x - 4)(x + 3) = 0
Setting each factor equal to zero:
x - 4 = 0 or x + 3 = 0
Solving for x, we get:
x = 4 or x = -3
Now, we substitute these values of x back into one of the original equations to find the corresponding values of y:
For x = 4, substituting into y = 3x - 4:
y = 3(4) - 4
y = 8
For x = -3, substituting into y = 3x - 4:
y = 3(-3) - 4
y = -13
Therefore, the solution to the linear-quadratic system y = x^2 - 4x + 8, y = 3x - 4 using the substitution method is (4, 8) and (-3, -13).