To solve the system of equations \( y = x^2 - 2x - 2 \) and \( y = 2x - 6 \) using the elimination method, we can set the equations equal to each other since they both equal \( y \).
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Set the equations equal to each other: \[ x^2 - 2x - 2 = 2x - 6 \]
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Rearrange the equation to one side: \[ x^2 - 2x - 2 - 2x + 6 = 0 \] Simplifying, \[ x^2 - 4x + 4 = 0 \]
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Factor the quadratic: \[ (x - 2)(x - 2) = 0 \] Thus, \[ x - 2 = 0 \implies x = 2 \]
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Now, substitute \( x = 2 \) back into either original equation to find \( y \). We'll use the linear equation: \[ y = 2(2) - 6 = 4 - 6 = -2 \] Therefore, we find one solution: \( (2, -2) \).
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Now, check if there are any other solutions by substituting \( y = 2x - 6 \) back into the quadratic equation: \[ 2x - 6 = x^2 - 2x - 2 \] Rearranging gives: \[ x^2 - 4x + 4 = 0 \] This confirms \( x = 2 \) is a double root, meaning it does not give another unique solution.
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To check for other solutions through substitution, reconsider the equations qualitatively or graphically. Since the quadratic function opens upwards, it can have no more intersections beyond the double root at \( (2, -2) \).
The complete verification confirms that the only solution to the system is:
\[ \boxed{(2, -2)} \]