$-5|2x+1|<10$, I keep getting the wrong answer I keep getting the wrong answer for $-5|2x+1|<10$.
The interval I come up with is $(-3/2, -1/2)$. The correct interval is $(-\infty,\infty)$.  I've figured out that if I simplify to just $|2x+1|>-2$, then all real numbers are true.
Why would I fully simplify other inequalities involving absolutes and not this one?  For example, simplifying $|2x-1|+7<13$, results in the interval $(-5/2,7/2)$ which is correct. Is this because of the division using $-5$? What am I missing?
EDIT: I'm sorry if I'm doing this completely wrong.  All of my work so far has been to completely simplify inequalities.  If this is the wrong approach, please let me know. My work:
$-5|2x+1|<10$
$|2x+1|>-2$
$2x+1>-2$ AND  $2x+1<2$
$2x+1>-2$
$2x>-3$
$x>-3/2$
AND
$2x+1<2$
$2x<1$
$x<-1/2$
Interval
$(-3/2,-1/2)$
 A: After dividing by $-5$, you get the following expression/inequality :
$$|2x+1| > -2$$
The inequality's "direction" changes when multiplying by a negative number $a$.
Note that the left hand side : $|2x+1| >0 \space \forall x \in \mathbb R$, since that yields from the definition of the absolute value. So you essentially want a positive value to be larger than a negative one, which is always true. That's why the correct answer is $x \in \mathbb R \Rightarrow x \in (-\infty, \infty)$.
Please show a complete elaboration regarding the initial answer though, as you may have made a mistake, so that we can point it out for you. 
Finally, the last example that you mentioned, is an absolute value being larger than a positive value, which has a specific answer/interval, since it does not hold for any $x$. This is a different case though from positive larger than negative, which always holds.
A: We have $$-5|2x+1|<10\implies |2x+1|>-2$$ which is true for all $x\in\mathbb{R}$ since the absolute value can never be less than $0$. Hence the interval.
A: The minus sign multiplies an absolute value, so the left side is at most $0$.  That means it will be true for all $x$.  When you got your interval you divided by $-5$ but did not reverse the inequality and dropped the sign to get to $|2x+1| \lt 2$.  Compare that to the inequality in your second paragraph and you will see the difference.  
The error in the work you show is when you go from $|2x+1| \gt -2$ to $2x+1 \gt -2$ AND $2x+1 \lt 2$.  If you want to strip off the absolute value signs you would have to say $(2x+1 \ge 0\ AND\ 2x+1 \gt -2)\ OR\ (2x+1 \lt 0\ AND\ 1-2x \gt -2)$  This will get you to $(-\infty,\infty)$ as well as the observation in the top paragraph.
