Questions about solving inequality: $2 < \frac{3x+1}{2x+4}$ Solve the inequality: $2 < \frac{3x+1}{2x+4}$  
Step 1: I simplified $\frac{3x+1}{2x+4}$ into: $3x+1-2x-4= x-3$.
Step 2: $2>x-3$ Here I subtracted $2$ from both sides into: $x>-5$ or $5<x$
However the answer in the book shows $-7>x>-2$. It's odd that $7-2=5$ but oke... 
So, I also tried using a sign diagram and with this I did get: $-2 for 2x+4$. I couldn't get $-7$ of of $(3x+1)$. It has probably something to do with the $2$ on the lhs. 
My questions are:
- Why is my first and second method wrong?
- How do I know if I should approach it using this method or with an sign diagram?   Sometimes the first method I use works, sometimes it doesn't. Please explain why.
- What other methods can I use to solve inequality's.
- Finally, please explain how I can solve this inequality.  

$$\frac{3x+1}{2x+4} >  2$$
$$\frac{3x+1}{2x+4} -  2 > 0$$
$$-\frac{x+7}{2(x+2)} > 0$$
Now equate the numerator and denominator to zero and draw this numbers on number line. 

Since sign of this equation is ">" we must determine intervals on which our function $f(x) = -\frac{x+7}{2(x+2)} $ takes positive values. To do this choose one number of each interval and substitute it to the function. For example, for number "-5": $f(-5) = \frac{1}{3}$ which is positive. For this equation it is one interval: $$ -7<x<-2$$
