Question

Solving Simultaneous Equations Using Elimination Quick Check
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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)
left parenthesis 2 comma negative 2 right parenthesis

(−2,−2) and (−2,2)

Answer 1

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

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

Since both expressions are equal to \( y \), we can set them equal to each other to eliminate \( y \):

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

Now, let's rearrange and simplify the equation:

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

This gives:

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

Now, substituting \( x = 2 \) back into one of the original equations to find \( y \). We'll use the linear equation:

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

Thus, we have one solution:

\[
(2, -2)
\]

Now, to determine if there are any other solutions, we note that the equation \( (x - 2)^2 = 0 \) indicates that the only solution for \( x \) is \( 2 \), which means there is only one pair \((2, -2)\).

The answer is:

\[
(2, -2)
\]

So, the correct response is:

**(2, -2)**