\[\frac{3x-1}{\| x+1 \|}\gt 2\]

.The denominator is always positive, and the numerator is positive for

\[3x-1 \gt 0 \rightarrow x \gt 1/3\]

.\[x+1 \gt 0 \]

if \[x \gt -1\]

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

\[3x-1 \gt 2(x+1)=2x+2 \rightarrow 3x-2x \gt 2+1 \rightarrow x \gt 3\]

.Suppose

\[\| \frac{3x-1}{ x+1} \| \gt 2\]

.Removing the modulus sign forces us to take signs into account. Either numerator and denominator have the same sign:

\[3x-1, \: x+1 \lt 0 \rightarrow x \lt 1/3, \: x \lt -1 \rightarrow x \lt -1\]

or \[3x-1, \: x+1 \gt 0 \rightarrow x \gt 1/3, \: x \gt -1 \rightarrow x \gt1/3\]

The numerator, denominator positive case is given above. Suppose numerator and denominator are both negative. Multiplying by a negative number reverses the inequality.

\[3x-1 \lt 2(x+1)=2x+2 \rightarrow 3x-2x \lt 2+1 \rightarrow x \lt 3,\]

But

\[x \lt -1\]

so \[x \gt 3 \]

or \[x \lt -1\]

(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,