Question

Use the elimination method to solve the linear-quadratic system y=x2+4x+5, y=12x−7.(1 point)
Responses

(2,17)
left parenthesis 2 comma 17 right parenthesis

(6,65) and (2,17)
left parenthesis 6 comma 65 right parenthesis and left parenthesis 2 comma 17 right parenthesis

(6,65)
left parenthesis 6 comma 65 right parenthesis

(−6,−79) and (−2,−31)

Answer 1

To solve the linear-quadratic system using the elimination method, we have the following equations:

1. \( y = x^2 + 4x + 5 \)
2. \( y = 12x - 7 \)

We can set the two equations equal to each other by substituting for \( y \):

\[
x^2 + 4x + 5 = 12x - 7
\]

Next, we move all terms to one side of the equation:

\[
x^2 + 4x + 5 - 12x + 7 = 0
\]

Simplifying this gives:

\[
x^2 - 8x + 12 = 0
\]

Now we can factor the quadratic:

\[
(x - 6)(x - 2) = 0
\]

So, the solutions for \( x \) are:

\[
x = 6 \quad \text{or} \quad x = 2
\]

Now we will find the corresponding \( y \)-values by substituting each \( x \) back into one of the original equations. Let’s use the linear equation \( y = 12x - 7 \).

1. For \( x = 6 \):

\[
y = 12(6) - 7 = 72 - 7 = 65
\]
So, one solution is \( (6, 65) \).

2. For \( x = 2 \):

\[
y = 12(2) - 7 = 24 - 7 = 17
\]
So, the other solution is \( (2, 17) \).

The solutions to the system are \( (6, 65) \) and \( (2, 17) \).

Thus, the final answer is:

**(6,65) and (2,17)**.