Use the elimination method to solve the linear-quadratic system y=x2−2x−2

Question

, y=2x−6
.(1 point)
Responses

(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)
left parenthesis negative 2 comma negative 10 right parenthesis

(2,−2)
and (−2,2)

Answer 1

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

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

To eliminate \( y \), we set the two expressions for \( y \) equal to each other:

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

Now, let's rearrange and combine like terms:

\[ x^2 - 2x - 2 - 2x + 6 = 0 \] \[ x^2 - 4x + 4 = 0 \]

Next, we can factor this equation:

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

This gives us:

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

Now that we have \( x = 2 \), we will substitute this value back into either original equation to find \( y \). We'll use the second equation:

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

So, we have one solution:

\[ (2, -2) \]

Next, we need to check if there are any other solutions. However, since the quadratic \( (x - 2)^2 = 0 \) has a double root at \( x = 2 \), we only have this one solution for \( x \), which gives us:

1 solution: \( (2, -2) \)

Thus, the solution to the system is:

\[ (2, -2) \]

So the correct response is:

(2, −2).