Use the elimination method to solve the linear-quadratic system y = x^2 - 2x -2, y=2x-6.
• (-2, -2) and (-2,2)
• (-2,-10)
• (2,-2)
• (2, -2) and (-2,2)
To solve the system using the elimination method, we need to eliminate one variable by manipulating one or both equations.
Given:
y = x^2 - 2x - 2 ...........(1)
y = 2x - 6 ....................(2)
We can start by subtracting equation (2) from equation (1) to eliminate y:
x^2 - 2x - 2 - (2x - 6) = 0
x^2 - 2x - 2 - 2x + 6 = 0
x^2 - 4x + 4 = 0
Now, we can factor this quadratic equation:
(x - 2)(x - 2) = 0
(x - 2)^2 = 0
We now have an equation with a double root of x = 2.
To find y, we substitute x = 2 into either equation (1) or (2):
Using equation (1):
y = (2)^2 - 2(2) - 2
y = 4 - 4 - 2
y = -2
Therefore, the solution to the system is (x, y) = (2, -2).
To check the given answer choices:
• (-2, -2) and (-2,2): These points do not satisfy both equations.
• (-2,-10): This point does not satisfy both equations.
• (2,-2): This point satisfies both equations.
• (2, -2) and (-2,2): These points satisfy both equations.
Therefore, the correct answer is (2, -2) and (-2, 2).