Question

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

Answer 1

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

1. Quadratic equation: \( y = x^2 - 2x - 2 \)
2. Linear equation: \( y = 2x - 6 \)

First, we can set the two equations equal to each other since they both equal \( y \):

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

Now, rearranging this equation gives:

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

This gives us a single solution for \( x \):

\[
x = 2
\]

Now, we can substitute \( x = 2 \) back into either equation to find \( y \). Let’s use the linear equation \( y = 2x - 6 \):

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

Thus, we have the solution:

\[
(x, y) = (2, -2)
\]

Now we can check if this solution also lies on the quadratic equation:

Substituting \( x = 2 \) into the quadratic equation:

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

Since both equations confirm that \( y = -2 \), the solution to the system of equations is:

\[
(2, -2)
\]

So the correct response is:

\((2, -2)\)