Inequalities Involution Absolutes

To solve inequalities involving a modulus, we have to be careful about removing the modulus sign, as there may be an unintended change of sign.
\[\frac{3x-1}{\| x+1 \|}\gt 2\]
The denominator is always positive, and the numerator is positive for  
\[3x-1 \gt 0 \rightarrow x \gt 1/3\]
\[x+1 \gt 0 \]
\[x \gt -1\]
  which is the case if  
\[x \gt 1/3\]
. We can write
\[\frac{3x-1}{ x+1 } \gt 2, \: x \gt 1/3\]

Numerator and denominator are both positive so  
\[3x-1 \gt 2(x+1)=2x+2 \rightarrow 3x-2x \gt 2+1 \rightarrow x \gt 3\]
\[\| \frac{3x-1}{ x+1} \| \gt 2\]
Removing the modulus sign forces us to take signs into account. Either numerator and denominator have the same sign:
\[3x-1, \: x+1 \lt 0 \rightarrow x \lt 1/3, \: x \lt -1 \rightarrow x \lt -1\]
\[3x-1, \: x+1 \gt 0 \rightarrow x \gt 1/3, \: x \gt -1 \rightarrow x \gt1/3\]

The numerator, denominator positive case is given above. Suppose numerator and denominator are both negative. Multiplying by a negative number reverses the inequality.
\[3x-1 \lt 2(x+1)=2x+2 \rightarrow 3x-2x \lt 2+1 \rightarrow x \lt 3,\]

\[x \lt -1\]
\[x \gt 3 \]
\[x \lt -1\]
Suppose only one of denominator and numerator are positive. It must be the numerator, since if the numerator were positive (
\[x \gt 1/3\]
) and the denominator negative (
\[x \lt -1\]
) which is not possible. Hence  
\[-1 \lt x \lt 1/3\]
  (2) .Hence
\[3x-1 \gt 2(x+1)=2x+2 \rightarrow 3x-2x \gt 2+1 \rightarrow x \gt 3,\]
This is inconsistent with (2) so only (1) is possible,

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