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If we have something like $|\frac{1}{x}|<3$ we now that this is the same as $|x|>\frac{1}{3}$ and we can write solutions as $x<-\frac{1}{3} \lor x> \frac{1}{3}$; but... if I try mecanically to view it as a system of equation like $$-3<\frac{1}{x}<3\iff\begin{cases}\frac{1}{x}<3\\\frac{1}{x}>-3\end{cases}$$ so working on every equation I have $$\begin{cases}x>\frac{1}{3}\\x<-\frac{1}{3}\end{cases}$$ that hasn't solutions... where is my error?

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Your first double inequality is an "and" inequality whereas your second double inequality is an "or" inequality.

  1. $$ -3<\frac{1}{x} \text{ and } \frac{1}{x}<3 $$

  2. $$ x<-\frac{1}{3} \text{ or } x>\frac{1}{3}$$

The or enters in the following way:

  1. $-3<\dfrac{1}{x}<3$ and since $x$ can be either positive or negative, then

  2. Either $-3<\dfrac{1}{x}<0$ or $0<\dfrac{1}{x}<3$ therefore

  3. Either $-3x>1>0$ or $0<1<3x$

  4. Either $x<-\dfrac{1}{3}$ or $\dfrac{1}{3}<x$

Notice that in steps $(2)$ and $(3)$ multiplication by a negative quantity reverses the sense of the first inequality.

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Obviously $x \in \mathbb{R} \setminus \{0\}$ hence $x \in (-\infty, 0) \cup (0, \infty)$. If $-3 < \frac{1}{x} < 3$ then if $x\in (0,\infty)$ then $-3 < \frac{1}{x}$ $\forall x \in (0,\infty)$. The other condition $\frac{1}{x} < 3$ yields $x > \frac{1}{3}$. Otherwise, if $x \in (-\infty, 0)$ then $\frac{1}{x} < 3$ $\forall x$ and the other condition is $-3 < \frac{1}{x} \Rightarrow 3 > -\frac{1}{x} \Rightarrow x < -\frac{1}{3}$ the sign changes again because $x < 0$. Hence $x \in (-\infty, -\frac{1}{3}) \cup (\frac{1}{3}, \infty)$

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i would consider two cases: $$x>0$$ and we have $$x>\frac{1}{3}$$ in the case of $$x<0$$ we get $$-\frac{1}{x}<3$$ which yields $$x<-\frac{1}{3}$$

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$\frac {1}{x}<3$ doesn't implie that $x>\frac {1}{3} $. that is your error.

to be true, $x $ must be positive.

so

$$(\frac {1}{x}<3 )\;\;\land \;\;(x>0) \;\implies x>\frac {1}{3} $$

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