Solving Inequalities: Cases and Solutions for \frac{3}{|x+1|-1}+\frac{2}{x}<1

[XJellieBX]

Problem:
$$\frac{3}{|x+1|-1}$$+$$\frac{2}{x}$$<1

My Solution:
There are 2 cases:
1) x$$\geq0$$,$$\frac{3}{x+1-1}$$+$$\frac{2}{x}$$<1
$$\frac{3}{x}$$+$$\frac{2}{x}$$<1
And you end up with... x>5.

2) x<0, $$\frac{3}{-x-1-1}$$+$$\frac{2}{x}$$<1
$$\frac{3x-2x-4}{-x^{2}-2x}$$<1
... a few reductions later...
$$x^{2}$$+3x-4>0
And the solution set for this case is, -4>x>1

My Question:
Is there something I'm missing or something else I need to do?

 

[Mentallic]

(edited: incorrect response)

 

[HallsofIvy]

"-4 > x> 1" implies -4> 1! x²+ 3x- 4= (x+ 4)(x- 1)> 0 if and only if the two factors are of the same sign. x+4> 0, x- 1> 0 give x> -4 and x> 1 which are both satisfied for x> 1. x+4< 0, x-1< 0 give x< -4, x< 1 which are both satisfied for x< -4.
x²+ 3x- 4> 0 for x< -4 OR x> 1.