Hi, good day

Can anyone helps me in solving this absolute inequality?

$\displaystyle

\frac{|x+1|}{|3x-5|}<1

$

Part 1:

$\displaystyle \frac{x+1}{3x-5}<1$

$\displaystyle \frac{x+1}{3x-5}-1<0$

$\displaystyle \frac{-2x+6}{3x-5}<0$

Part 2:

$\displaystyle \frac{x+1}{3x-5}>-1$

$\displaystyle \frac{x+1}{3x-5}+1>0$

$\displaystyle \frac{4x-4}{3x-5}>0$

Next?

p/s:I know there is other method, something like squaring both sides, but I find it untrue for certain questions.