Question

Question
Use the elimination method to solve the linear-quadratic system y=x2−2x−2
y
=
x
2
−
2
x
−
2
, y=2x−6
y
=
2
x
−
6
.(1 point)
Responses

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

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

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

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

Answer 1

To solve the system of equations using the elimination method, we start with the given equations:

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

We will set the two expressions for \( y \) equal to each other:

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

Now, we will rearrange this equation to form a standard quadratic equation by bringing all terms to one side:

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

This simplifies to:

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

Next, we can factor the quadratic:

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

This gives us a double root at:

\[
x = 2
\]

Now we plug \( x = 2 \) back into Equation 2 to find the corresponding \( y \):

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

So we have found one solution:

\[
(2, -2)
\]

Now, we need to check if there are other solutions. We can verify if there are double solutions by checking the quadratic again:

Setting the quadratic equal to zero gives a single solution \( x = 2 \) with a double root, which suggests there are no other x-values producing another solution.

Thus, the only solution to the system of equations is:

\[
\boxed{(2, -2)}
\]