Question Which of the following uses the substitution method to solve linear-quadratic system y=x2−4x+8 , y=3x−4 ?(1 point) Responses (4,8) left parenthesis 4 comma 8 right parenthesis (−4,−16) and (−3,−13) left parenthesis negative 4 comma negative 16 right parenthesis and left parenthesis negative 3 comma negative 13 right parenthesis (4,8) and (3,5) left parenthesis 4 comma 8 right parenthesis and left parenthesis 3 comma 5 right parenthesis (3,5)

Question

Question Which of the following uses the substitution method to solve linear-quadratic system y=x2−4x+8 , y=3x−4 ?(1 point) Responses (4,8) left parenthesis 4 comma 8 right parenthesis (−4,−16) and (−3,−13) left parenthesis negative 4 comma negative 16 right parenthesis and left parenthesis negative 3 comma negative 13 right parenthesis (4,8) and (3,5) left parenthesis 4 comma 8 right parenthesis and left parenthesis 3 comma 5 right parenthesis (3,5)

Answer 1

To solve the linear-quadratic system using the substitution method, we need to substitute the expression for y from one equation into the other equation and solve for x.

First, we substitute y = 3x - 4 into the equation y = x^2 - 4x + 8:

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

Next, we rearrange the equation to have all terms on one side:

x^2 - 4x - 3x + 8 + 4 = 0.

x^2 - 7x + 12 = 0.

Now, we factor the quadratic equation:

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

Setting each factor equal to zero:

x - 3 = 0 and x - 4 = 0.

x = 3 and x = 4.

Substituting these values of x back into the equation y = 3x - 4:

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

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

Therefore, the solution to the linear-quadratic system is (3, 5) and (4, 8).