Case I : x > 2

1 ≥ 0

It is true that 1 is always greater than 0 so case I is also true x > 2

Case II : x < 2

-1 ≥ 0

It is false that -1 is not greater than 0 so case II is also false.

So, the final solution is x > 2 i.e., x ∈ (2, ∞)