Question

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,−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 system of equations using the elimination method, we have the following two equations:

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

We can eliminate \( y \) by setting these two equations equal to each other:

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

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

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

Now, we can factor this quadratic:

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

This gives us a double root:

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

Next, we substitute \( x = 2 \) back into one of the original equations to find \( y \). We can use the linear equation:

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

So, we have the point \( (2, -2) \).

To check if there are any other solutions, we can substitute \( x = 2 \) into the quadratic equation:

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

Both equations give us the same point \( (2, -2) \).

Now, let's check if the other potential points given in the answer options fit:

1. \((-2, -2)\)
2. \((-2, 2)\)
3. \((-2, -10)\)

Since we only found one solution so far, we check \( x = -2 \):

For the linear equation:
\[
y = 2(-2) - 6 = -4 - 6 = -10
\]

And for the quadratic:
\[
y = (-2)^2 - 2(-2) - 2 = 4 + 4 - 2 = 6
\]

Since \( (-2, -10) \) does not match \( (-2, 6) \), this point does not work. The same would happen for \((-2, -2)\) and \((-2, 2)\).

The only solution to the system is:

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