Inequalities Question Regarding the Work I Did to Solve it

The set of all non-zero real numbers x such that 6/x > 2 can be expressed as an open interval of the form (a, b). Express a and b as integers.

Now I did the following...

Case I:
x > 0
3 > x

Case II:
x < 0
3 < x

Since the first and second statement contradict each other in Case II, Case I is correct. Therefore, the interval is (0, 3), so a = 0, b = 3. I knew logically that a = 0 because if a < 0 then 6/x < 0, so the condition 6/x < 2 would not be satisfied, but how would I know this based on my work from Case I?

___________________________

"but how would I know this based on my work from Case I?"
As you have noticed, x cannot be negative. So, Case II never gets to determine the possibility x>3.

Consider this:
Given 6/x > 2:
Because we cannot divide by 0, we have x≠0.
And, because 6/x is greater than 2, it follows
that 6/x is positive so that x must be positive.

And, for x > 0:
    6/x > 2
      6 > 2x
      3 > x

  Therefore,
   0 < x < 3
        or
Open Interval (0,3)


Have a good one!!
__________________________