Question

Use the elimination method to solve the linear-quadratic system y=x2−2x−2
, y=2x−6
.(1 point)
Responses

(2,−2)
and (−2,2)
left parenthesis 2 comma negative 2 right parenthesis and left parenthesis negative 2 comma 2 right parenthesis

(2,−2)
left parenthesis 2 comma negative 2 right parenthesis

(−2,−2)
and (−2,2)
left parenthesis negative 2 comma negative 2 right parenthesis and left parenthesis negative 2 comma 2 right parenthesis

(−2,−10)

Answer 1

To solve the given system of equations using the elimination method, we have:

1. \( y = x^2 - 2x - 2 \)
2. \( y = 2x - 6 \)

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

\[
x^2 - 2x - 2 = 2x - 6
\]

Now, let's move all terms to one side of the equation:

\[
x^2 - 2x - 2 - 2x + 6 = 0
\]

This simplifies to:

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

Factoring the quadratic:

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

This gives us:

\[
x - 2 = 0 \quad \Rightarrow \quad x = 2
\]

Now that we have \( x = 2 \), we can substitute it back into one of the original equations to find \( y \). We'll use the second equation for this:

\[
y = 2(2) - 6 = 4 - 6 = -2
\]

So, we found one solution: \( (2, -2) \).

Next, since the quadratic equation \( (x - 2)^2 = 0 \) has a double root for \( x = 2 \), we don't have any other \( x \) values to find. Now, let’s check if there are any other solutions by substituting back into the first equation:

For \( y = x^2 - 2x - 2 \):
\[
y = (2)^2 - 2(2) - 2 = 4 - 4 - 2 = -2
\]

This confirms that our point \( (2, -2) \) is indeed a solution.

Therefore, the solution to the given system of equations is:

**(2, −2)**