Use the elimination method to solve the linear-quadratic system y = x ^ 2 - 2x - 2 2y y = 2x - 6 point) (- 2, - 10); (2, - 2); (2, - 2) and (- 2, 2); (- 2, - 2) and (- 2, 2)
First, we need to set up the system of equations using the given points.
For the point (-2, -10), we have:
y = x^2 - 2x - 2
-10 = (-2)^2 - 2(-2) - 2
-10 = 4 + 4 - 2
-10 = 6
This equation is not true, so the point (-2, -10) is not a solution to the system.
For the point (2, -2), we have:
y = x^2 - 2x - 2
-2 = (2)^2 - 2(2) - 2
-2 = 4 - 4 - 2
-2 = -2
This equation is true, so the point (2, -2) is a solution to the system.
For the point (2, -2) and (-2, 2), we have:
y = x^2 - 2x - 2
-2 = (-2)^2 - 2(-2) - 2
-2 = 4 + 4 - 2
-2 = 6
This equation is not true, so the points (2, -2) and (-2, 2) are not solutions to the system.
Therefore, the only solution to the system is (2, -2).
RU sure
Apologies for the mistake. Let's verify the solution again.
For the point (-2, -10), we have:
y = x^2 - 2x - 2
-10 = (-2)^2 - 2(-2) - 2
-10 = 4 + 4 - 2
-10 = 6
This equation is not true, so the point (-2, -10) is not a solution to the system.
For the point (2, -2), we have:
y = x^2 - 2x - 2
-2 = (2)^2 - 2(2) - 2
-2 = 4 - 4 - 2
-2 = -2
This equation is true, so the point (2, -2) is a solution to the system.
For the point (-2, 2), we have:
y = x^2 - 2x - 2
2 = (-2)^2 - 2(-2) - 2
2 = 4 + 4 - 2
2 = 6
This equation is not true, so the point (-2, 2) is not a solution to the system.
Therefore, the only solution to the system is (2, -2).