Use the elimination method to solve the linear-quadratic system y=x2−2x−2

Question

, y=2x−6
.(1 point)
Responses

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

(−2,−10)
left parenthesis negative 2 comma negative 10 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)

Answer 1

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

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

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

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

Next, we will move all terms to one side of the equation:

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

Combining like terms gives us:

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

This is a perfect square trinomial that can be factored:

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

Setting the factor equal to zero, we solve for \( x \):

\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]

Now that we have \( x = 2 \), we can substitute this value back into one of the original equations to find \( y \). We'll use the second equation:

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

So, we have the solution:

\[ (x, y) = (2, -2) \]

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

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

Thus, the only solution for the system is \((2, -2)\).

The final answer is:

(2, -2)