Question

Use the elimination method to solve the linear-quadratic system y=x2−2x−2, y=2x−6.(1 point)%0D%0AResponses%0D%0A%0D%0A(2,−2)%0D%0Aleft parenthesis 2 comma negative 2 right parenthesis%0D%0A%0D%0A(−2,−2) and (−2,2)%0D%0A left parenthesis negative 2 comma negative 2 right parenthesis and left parenthesis negative 2 comma 2 right parenthesis%0D%0A%0D%0A(2,−2) and (−2,2)%0D%0Aleft parenthesis 2 comma negative 2 right parenthesis and left parenthesis negative 2 comma 2 right parenthesis%0D%0A%0D%0A(−2,−10)

Answer 1

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

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

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:

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

Next, factor the quadratic:

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

This implies:

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

Now we will substitute \( x = 2 \) into the linear equation to find \( y \):

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

Thus, we have one solution:

\[
(2, -2)
\]

Now, let's check if there are any other solutions by substituting \( y = 2x - 6 \) into the quadratic equation again. Since we found \( (2, -2) \), we can check for a possible repeat:

1. Substitute \( y \) into the quadratic:
\( 2x - 6 = x^2 - 2x - 2 \)
- Rearranging gives us:
\[
x^2 - 4x + 4 = 0 \implies (x - 2)^2 = 0
\]

We already established that the quadratic equation has a double root at \( x = 2 \) with \( y = -2 \). This means the only solution for the system is:

\[
(2, -2)
\]

Hence, the answer is:

\[
\text{(2, −2)}
\]