Question

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

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

(2,−2)
and (−2,2)
left parenthesis 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

(−2,−10)

Answer 1

To solve the linear-quadratic system using the elimination method, we'll substitute the linear equation into the quadratic equation.

1. Start with the equations:
\[
y = x^2 - 2x - 2 \quad \text{(1)}
\]
\[
y = 2x - 6 \quad \text{(2)}
\]

2. Substitute equation (2) into equation (1):
\[
2x - 6 = x^2 - 2x - 2
\]

3. Rearrange the equation:
\[
0 = x^2 - 2x - 2 - (2x - 6)
\]
This simplifies to:
\[
0 = x^2 - 4x + 4
\]

4. Factor the quadratic equation:
\[
0 = (x - 2)^2
\]

5. Solve for \(x\):
\[
x - 2 = 0 \implies x = 2
\]

6. Substitute \(x = 2\) back into equation (2) to find \(y\):
\[
y = 2(2) - 6 = 4 - 6 = -2
\]

7. We have found one solution: \((2, -2)\).

Now, let's check if there are other possible solutions by substituting \(y\) back into both original equations. The quadratic equation has already been fully factored, indicating that \((2, -2)\) is the repeated root.

The system yields one solution:
The correct answer is:
\[
\text{(2, -2)}
\]