# Solve for $x$: $3+|x-9|<\frac{2|x-1|}{x}$

I moved one part of the inequality to the other to create the following:

$$\frac{2|x-1|}{x}-3-|x-9|>0$$

Eventually, I get to a case where I have 2 inequalities after opening 1 of the absolute values, but then get confused on what do I do with the second one. Am I supposed to get 4 inequalities? Isn't there a simpler method which I am missing?

Unfortunately, there is no easier way than splitting this into different cases. You can restrict yourself to the cases $$0 < x \leq 1$$, $$1 < x < 9$$ and $$x \geq 9$$. However, two of these cases are immediately impossible, so you only end up with one case. Can you finish?