# Thread: Inequality - x on denominator

1. A fraction is positive in 2 conditions : a) Numerator and denominator both positive or b) Numerator and denominator both negative.
If one is positive and the other is negative then the fraction would be negative.

In your fraction numerator is a standard positive number, so it would be always positive. That means you are in condition (a) above so you need denominator to be positive.

Originally Posted by Stroodle
Using the Abstractionists method wouldn't it be $\displaystyle 3x+2<6$ and $\displaystyle 3x+2>0$ ??
Why it should be $\displaystyle 3x+2<6$?? We talk about the fraction $\displaystyle \frac{12}{3x+2}>0$ yes? Here you just need to check $\displaystyle 3x+2>0$ and you have your solution.
Hope i said it well, because English isn't my mother language

2. Thanks Sentinel, but I meant it for $\displaystyle \frac{12}{3x+2}>2$

I should've been clearer

3. For $\displaystyle \frac{12}{3x+2}>2 \Rightarrow \frac{12}{3x+2}-2>0 \Rightarrow \frac{8-6x}{3x+2}>0$ and you take :

i) 8-6x>0 and 3x+2>0
ii) 8-6x<0 and 3x+2<0

You solve i) and ii) and you have for which x it stands...

Page 2 of 2 First 12