<s>[INST] Why does $-1<\frac{7}{x+4}$ imply $x<-11$? I keep getting $x>-11$ I'm trying to solve for the $|\frac{7}{x+4}|<1$. Which becomes $-1<\frac{7}{x+4}<1$.
Evaluating the positive side is fine, $3<x,$ but for the negative side:
$-1<\frac{7}{x+4}$ should evaluate to $-11 > x$. But I keep getting $-11 < x.$
My working:
$$-1<\frac{7}{x+4}\\
-1(x+4)<7\\-x < 11\\
x>-11.$$
This results in the answer $x>-11,  x>3,$ which doesn't make sense.
 [/INST] Edited for clarity.

Expanding on the comment of sonicsid, when you are faced with the necessity of clearing the denominator of $(x + 4)$, since $x + 4 = 0$ is disallowed, you must break the analysis into two cases:
Case 1: $(x + 4) > 0.$ 
$-1 < \frac{7}{x+4} < 1.$ 
$ (-1) (x+4) < 7 < (x+4).$ 
$(x + 4) > -7$ and $(x + 4) > 7.$
This means that all three of the following constraints must be satisfied for Case 1:

*

*$x > -4.$

*$x > -11.$

*$x > 3.$
Therefore, the Case 1 solutions resolve to 
$x > 3.$

Case 2: $(x + 4) < 0.$ 
$-1 < \frac{7}{x+4} < 1.$ 
$(-1) (x + 4) > 7$ and $7 > (x + 4).$
This means that all three of the following constraints must be satisfied for Case 2:

*

*$x < -4.$

*$x < -11.$

*$x < 3.$
Therefore, the Case 2 solutions resolve to 
$x < -11.$

Putting Case 1 and Case 2 together:
Either $x > 3$ or $x < -11.$
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