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> Problem: $\dfrac{2}{|x-4|}>1$ Express the solutions using intervals
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My attempt using the Definition of Modulus:
$$\dfrac{2-|x-4|}{|x-4|}>0$$

$$CASE A:x-4\ge 0\Rightarrow x\ge4\Rightarrow x\in [4,\infty) \Longrightarrow |x-4|=x-4 $$ 
$\dfrac{6-x}{x-4}>0$
$$\Longrightarrow x\in (4,6) \cap [4,\infty)\Longrightarrow x\in (4,6)$$
$$CASE B:x-4< 0\Rightarrow x<4\Rightarrow x\in (-\infty,4) \Longrightarrow |x-4|=-(x-4) $$ 
$\dfrac{x-2}{x-4}>0$
$$\Longrightarrow x\in (-\infty,4)\cap[(-\infty,2)\cup(4,\infty)]$$
$$\Rightarrow \in(-\infty,2)$$
Thus, the final solution through this method should be $x\in (4,6)\cup(-\infty,2)$  

However, we know that the answer is $x\in(2,6)-${4}
Could somebody please tell me where I've gone wrong? Thanks so much in advance!