To solve the linear-quadratic system using the elimination method, we have the following equations:
- \( y = x^2 - 2x - 2 \) (quadratic equation)
- \( y = 2x - 6 \) (linear equation)
Since both equations are equal to \( y \), we can set them equal to each other:
\[ x^2 - 2x - 2 = 2x - 6 \]
Now, let's move all terms to one side of the equation:
\[ x^2 - 2x - 2 - 2x + 6 = 0 \]
This simplifies to:
\[ x^2 - 4x + 4 = 0 \]
Next, factor the quadratic:
\[ (x - 2)^2 = 0 \]
This implies:
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]
Now we will substitute \( x = 2 \) into the linear equation to find \( y \):
\[ y = 2(2) - 6 = 4 - 6 = -2 \]
Thus, we have one solution:
\[ (2, -2) \]
Now, let's check if there are any other solutions by substituting \( y = 2x - 6 \) into the quadratic equation again. Since we found \( (2, -2) \), we can check for a possible repeat:
- Substitute \( y \) into the quadratic:
\( 2x - 6 = x^2 - 2x - 2 \)
- Rearranging gives us: \[ x^2 - 4x + 4 = 0 \implies (x - 2)^2 = 0 \]
We already established that the quadratic equation has a double root at \( x = 2 \) with \( y = -2 \). This means the only solution for the system is:
\[ (2, -2) \]
Hence, the answer is:
\[ \text{(2, −2)} \]