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

(2,−2)

Answer 1

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

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

We'll start by substituting Equation 2 into Equation 1.

Setting these equations equal to each other, we have:

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

Now, let's rearrange the equation to one side:

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

Combine like terms:

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

This simplifies to:

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

Setting the equation to zero gives:

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

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

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

So one solution is \( (2, -2) \).

Next, we will check for any other possible solutions by solving the quadratic equation. Since \((x - 2)^2 = 0\) only gives us \( x = 2 \) as a solution, we will not have additional \(x\) values from this equation.

Now let's confirm by substituting \( x = 2 \) back into Equation 1:

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

This confirms the solution \( (2, -2) \).

Since this quadratic produced only one unique solution, there are no other intersections.

Therefore, the solution set for the given linear-quadratic system is:

**(2, −2)**