Use the elimination method to solve the linear-quadratic system: y = x^2 - 2x - 2, y = 2x - 6

Answer 1

To solve the given linear-quadratic system using the elimination method, we have the following equations:

1. \( y = x^2 - 2x - 2 \) (quadratic equation)
2. \( y = 2x - 6 \) (linear equation)

We can eliminate \( y \) by setting the two equations equal to each other:

\[ x^2 - 2x - 2 = 2x - 6 \]

Next, we will rearrange this equation to set it to zero:

1. Move \( 2x \) and \( -6 \) to the left side:

\[ x^2 - 2x - 2 - 2x + 6 = 0 \]

2. Combine like terms:

\[ x^2 - 4x + 4 = 0 \]

Now, we can factor the quadratic equation:

\[ (x - 2)^2 = 0 \]

This gives us:

\[ x - 2 = 0 \implies x = 2 \]

Next, we can substitute back to find the corresponding \( y \) value. We can use the linear equation \( y = 2x - 6 \):

\[ y = 2(2) - 6 = 4 - 6 = -2 \]

Thus, the solution to the system of equations is \( (x, y) = (2, -2) \).

To summarize, the final solution is:

\[ \boxed{(2, -2)} \]