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Problem: $\dfrac{2}{|x-4|}>1$ Express the solutions using intervals

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!

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  • $\begingroup$ Actually the answer is $x \in (2,4) \cup (4,6)$ since the left hand side is not defined at $x = 4$. $\endgroup$ – MathNewbie May 24 '15 at 7:34
  • $\begingroup$ Oh yes, sorry for the error. $\endgroup$ – Ishan May 24 '15 at 7:37
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    $\begingroup$ the idea is that in case B , when you have $|x-4|=-(x-4)$ you have to pay attention also to the denomenator , so you get $$ \frac{2-x}{x-4}>0$$ So I suggest writting $$ \frac{x-2}{4-x}>0$$ this implies (as $4-x >0$) $$ x-2 >0$$ then $x \in (2,4)$. Thus for the final answer $x \in (2,6) -\{ 4 \}$ $\endgroup$ – Nizar May 24 '15 at 7:37
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    $\begingroup$ Oh... Sorry, hadn't noticed that I hadn't changed the denominator. Thanks so much for pointing out my error! $\endgroup$ – Ishan May 24 '15 at 7:40
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In case B) when you substitute |x-4|=-(x-4) then this should imply (x-2)/(4-x) not (x-2)/(x-4). Check it out

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  • $\begingroup$ Yes, Sir, I forgot to do that. Thanks a lot for your help! $\endgroup$ – Ishan May 24 '15 at 8:22
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Multiply by $|x-4|$ and then read the inequality out loud: “The distance of $x$ to $4$ is less than $2$.” And be aware of that $x=4$ is not a solution.

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  • $\begingroup$ Sir, I know how to solve by multiplying both sides by |x-4| (since it is always positive, hence direction of the inequality remains the same). But sir, why is my method yielding a wrong answer? $\endgroup$ – Ishan May 24 '15 at 7:36
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You don't have to consider separate cases. Since $|x-4|$ is not negative you can multiply across without changing the inequality.

You are then solving $|x-4|<2\Rightarrow-2<x-4<2$. Then you get the solution set you are supposed to get.

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