To solve the system of equations using the elimination method, we start with the two equations:
- \( y = x^2 - 2x - 2 \) (quadratic equation)
- \( y = 2x - 6 \) (linear equation)
Since the elimination method typically applies to linear equations, we can substitute the expression for \( y \) from the linear equation into the quadratic equation.
Now, set the second equation equal to the first:
\[ 2x - 6 = x^2 - 2x - 2 \]
Next, we will rearrange this equation to form a standard quadratic equation:
\[ 0 = x^2 - 2x - 2 - (2x - 6) \] \[ 0 = x^2 - 2x - 2 - 2x + 6 \] \[ 0 = x^2 - 4x + 4 \]
Now, we can factor this quadratic:
\[ 0 = (x - 2)(x - 2) \]
Thus, we have a double root:
\[ x - 2 = 0 \Rightarrow x = 2 \]
Next, we can find the corresponding \( y \) value using the linear equation:
\[ y = 2(2) - 6 = 4 - 6 = -2 \]
So, the solution to the system is:
\[ (2, -2) \]
Now, let's check our answer. We can plug \( x = 2 \) back into the quadratic equation:
\[ y = 2^2 - 2(2) - 2 = 4 - 4 - 2 = -2 \]
This confirms that both equations yield the same point \((2, -2)\).
Thus, the correct response is:
\[ (2, -2) \]