Is it allowed to solve this inequality $x|x-1|>-3$ by dividing each member with $x$? Is it allowed to solve this inequality  $x|x-1|>-3$ by dividing each member with $x$? What if $x$ is negative?
My textbook provides the following solution:

Divide both sides by $x: $ $\frac { x | x - 1 | } { x } > \frac { - 3
 } { x } ; \quad x \neq 0$
Simplify: $| x - 1 | > - \frac { 3 } { x } ; \quad x \neq 0$

Edit: provided textbook's solution

For $x\geq0$ this inequality is always true. 
Assume that $x<0$, so $x=-y$ for some positive $y$ and we get $$y|\;\underbrace{y+1}_{>0}\;|<3\implies y(y+1)<3 ...$$
