So you have 3 < |x-5|
First of all you must define the module function:
|x-5|= x-5 , for x-5>=0, meaning for x>=5
|x-5|= 5-x , for x-5<0, meaning for x<5
So we go on with the inequality on two sides:
We take first x>=5. We have 3<x-5 =>x>3+5=>x>8. And because x is supposed to be greater than 5, we have the final partial solution: x>8.
Next, we have x<5. Our inequality becomes: 3<5-x=>x<5-3=>x<2. Because x<5, the final second partial solution is x<2.
This is the last result.