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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 negative 2 comma negative 2 right parenthesis and left parenthesis negative 2 comma 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)
left parenthesis 2 comma negative 2 right parenthesis

(−2,−10)

1 answer

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

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

We'll set the expressions for \( y \) equal to each other:

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

Now, we will rearrange the equation to bring all terms to one side:

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

This simplifies to:

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

Factoring the quadratic:

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

So, we have:

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

Next, we substitute \( x = 2 \) back into either equation to find \( y \). We'll use the linear equation:

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

Thus, we have one solution:

\[ (2, -2) \]

Now we check if there are any other solutions. The equation \( (x - 2)^2 = 0 \) indicates that the solution \( x = 2 \) has a multiplicity of 2, meaning this is the only solution for \( x \).

Thus, the solution to the system is:

\[ (2, -2) \]

The correct answer is:

(2,−2).