Inequality - x on denominator

Hi all, I've got a inequality question. Should be basic but it's stuffing me up for some reason. I'll explain why in a moment. $\displaystyle \frac{12}{3x + 2} > 0 $ I'll go through a sample question, to explain how I usually solve it. $\displaystyle \frac{12}{3x + 2} > 2 \frac{12}{3x + 2} - 2 > 0 \frac{8 - 6x}{3x + 2} > 0 $ Then solve for each case... $\displaystyle 8 - 6x = 0 x = \frac{4}{3} 3x + 2 = 0 x = \frac{-2}{3} $ Then plot these on a number line and determine the range etc. What stuffs me up with the original question is the > 0. I've got nothing on the other side, to bring over and essentially get the x's up the top. The correct answer is -2/3 < x < 1/3. I can get the first half, however I'm not sure how to get the 1/3 answer. P.s. Sorry but the formatting seems to be stuffing up for some reason. I had this nicely spaced out but posting seemed to destroy it.