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)
and (3,5)
left parenthesis 4 comma 8 right parenthesis and left parenthesis 3 comma 5 right parenthesis

(4,8)
left parenthesis 4 comma 8 right parenthesis

(3,5)
left parenthesis 3 comma 5 right parenthesis

(−4,−16)
and (−3,−13)

Answer 1

To solve the linear-quadratic system of equations given by:

1. \( y = x^2 - 4x + 8 \) (quadratic)
2. \( y = 3x - 4 \) (linear)

we can use the substitution method.

Since both equations equal \( y \), we can set them equal to each other:

\[
x^2 - 4x + 8 = 3x - 4
\]

Now, rearranging the equation gives:

\[
x^2 - 4x - 3x + 8 + 4 = 0
\]

\[
x^2 - 7x + 12 = 0
\]

Next, we can factor the quadratic:

\[
(x - 3)(x - 4) = 0
\]

Setting each factor to zero gives us the possible \( x \) values:

\[
x - 3 = 0 \quad \Rightarrow \quad x = 3
\]
\[
x - 4 = 0 \quad \Rightarrow \quad x = 4
\]

Next, we substitute these \( x \) values back into either of the original equations to find the corresponding \( y \) values. We'll use the linear equation \( y = 3x - 4 \).

1. For \( x = 3 \):

\[
y = 3(3) - 4 = 9 - 4 = 5 \quad \Rightarrow \quad (3, 5)
\]

2. For \( x = 4 \):

\[
y = 3(4) - 4 = 12 - 4 = 8 \quad \Rightarrow \quad (4, 8)
\]

Thus, the solutions to the system are:

\[
(3, 5) \quad \text{and} \quad (4, 8)
\]

From your options, the correct response is:

**(4,8) and (3,5)**.