To solve the given system of equations using the elimination method, we have:
- \( y = x^2 - 2x - 2 \)
- \( y = 2x - 6 \)
Since both equations are equal to \( y \), we can set them equal to each other:
\[ x^2 - 2x - 2 = 2x - 6 \]
Now, let's move all terms to one side of the equation:
\[ x^2 - 2x - 2 - 2x + 6 = 0 \]
This simplifies to:
\[ x^2 - 4x + 4 = 0 \]
Factoring the quadratic:
\[ (x - 2)(x - 2) = 0 \]
This gives us:
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \]
Now that we have \( x = 2 \), we can substitute it back into one of the original equations to find \( y \). We'll use the second equation for this:
\[ y = 2(2) - 6 = 4 - 6 = -2 \]
So, we found one solution: \( (2, -2) \).
Next, since the quadratic equation \( (x - 2)^2 = 0 \) has a double root for \( x = 2 \), we don't have any other \( x \) values to find. Now, let’s check if there are any other solutions by substituting back into the first equation:
For \( y = x^2 - 2x - 2 \): \[ y = (2)^2 - 2(2) - 2 = 4 - 4 - 2 = -2 \]
This confirms that our point \( (2, -2) \) is indeed a solution.
Therefore, the solution to the given system of equations is:
(2, −2)