To solve the system of equations using the elimination method, we first rewrite the system clearly:
- \( y = x^2 - 2x - 2 \)
- \( y = 2x - 6 \)
We can set these two equations equal to each other since they both equal \(y\):
\[ x^2 - 2x - 2 = 2x - 6 \]
Next, we'll move all terms to one side to form a standard quadratic equation:
\[ x^2 - 2x - 2 - 2x + 6 = 0 \]
Combine like terms:
\[ x^2 - 4x + 4 = 0 \]
Next, we can factor the quadratic:
\[ (x - 2)(x - 2) = 0 \]
Thus, we have:
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]
Since this is a repeated root, we only have one value for \(x\) (i.e., \(x = 2\)).
Now we will find the corresponding \(y\) value by substituting \(x = 2\) back into either of the original equations. We'll use the second equation for simplicity:
\[ y = 2(2) - 6 \] \[ y = 4 - 6 \] \[ y = -2 \]
Thus, one solution to the system is \((2, -2)\).
Next, we need to check if there are any other solutions. Since we have a quadratic equation, we can see if it may yield any other values by substituting our \(x = 2\) back into the first equation to verify if there might be additional solutions.
Going back to the first equation:
\[ y = x^2 - 2x - 2, \]
At \(x = 2\):
\[ y = 2^2 - 2(2) - 2 \] \[ y = 4 - 4 - 2 \] \[ y = -2, \]
Verifying for any additional values:
The quadratic equation \(x^2 - 4x + 4 = 0\) only produces one unique solution \(x=2\) which means that the original system only has one solution at \((2, -2)\).
Now, let's check if by symmetry or the nature of the quadratic could yield surfaces through the concept of equal values:
Since we derived it from quadratic roots, recap from primary values of y via substitution of original equations leads us nowhere else.
Thus the final solution set for the given system is:
\[ \text{(2, -2)} \]
Hence the correct answer from your given choices is:
\((2, -2)\).
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