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 negative 2 comma negative 2 right parenthesis and left parenthesis negative 2 comma 2 right parenthesis

(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,−10)

Answer 1

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

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

We'll set the expressions for \( y \) equal to each other:

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

Now, we will rearrange the equation to bring all terms to one side:

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

This simplifies to:

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

Factoring the quadratic:

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

So, we have:

\[
x - 2 = 0 \implies x = 2
\]

Next, we substitute \( x = 2 \) back into either equation to find \( y \). We'll use the linear equation:

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

Thus, we have one solution:

\[
(2, -2)
\]

Now we check if there are any other solutions. The equation \( (x - 2)^2 = 0 \) indicates that the solution \( x = 2 \) has a multiplicity of 2, meaning this is the only solution for \( x \).

Thus, the solution to the system is:

\[
(2, -2)
\]

The correct answer is:

**(2,−2)**.