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,−10)
left parenthesis negative 2 comma negative 10 right parenthesis

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

(−2,−2) and (−2,2)

Answer 1

To solve the system of equations \( y = x^2 - 2x - 2 \) and \( y = 2x - 6 \) using the elimination method, we will set the two expressions for \( y \) equal to each other:

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

Next, we rearrange this equation to set it to zero:

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

Now, we can factor the quadratic:

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

This gives us the solution:

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

Next, we substitute \( x = 2 \) back into either equation to find \( y \). Using \( y = 2x - 6 \):

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

So, one solution to the system is \( (2, -2) \).

Now, we can check whether there are other solutions by evaluating other possibilities. Since the quadratic equation we derived was a perfect square, it means there is a double root, which is \( x = 2 \). Thus, the only solution for \( (x, y) \) in this system is:

\[
(2, -2)
\]

Therefore, the answer is:

**(2,−2)**