Question 18407
 When you think about the x-axis, just imagine that y=0. Of course, y can be
anything if you are only talking about the x-axis, which is horizontal.
  With y=0, you are always on the x-axis itself.

  There are only (3) things that x can be:

      (1) negative
      (2) zero
      (3) positive

  So now you're on the x-axis, and you wan to know where 2x > 0 exists and where
it doesn't exist. You know (or should know) that zero divided by anything
except zero is still zero.

      x > 0/2    --divide both sides of inequality by 2
      x > 0      --what I said last paragraph

  Look at the three things that x can be. The only one that fits x > 0 is
(3) positive. Now the trick is to let y be anything it wants to be. That
means zero, negative, or positive.  Your solution is that x can be anywhere
in the 1st and 4th quadrants, letting y roam free.


                             2nd|1st
                                   
                             ---+---
                             3rd|4th
                                   
  Remember that x can't be zero, that would mean x = 0, not x > 0.
Do you want to shade everywhere that you don't find a solution to
x > 0  ?  Then you must shade the 2nd and 3rd quadrants and also
shade the y-axis- because the y-axis is where x=0 and y can roam free.

 Hope this helps. I've gotten confused numerous times, too, and the
best advice I have is to give it the effort it requires until you're
confident. Good luck.