<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] Just to add to the noise, this is how I would personally attack this problem.  This is a matter of personal style and may not work for you.
$|\frac 7{x+4}| < 1$.
Either $\frac 7{x+4}$ is non-negative (positive or zero) or it is negative.
It is positive if $x+4$ is positive.  It is negative if $x+4$ is negative.  (And it is never $0$ as $\frac ab \ne 0; a=0$ and we can't have $x+4 = 0$ as we can't have $0$ in a denominator).
Case 1:
If $\frac 7{x+4}$ is positive then $x+4$ is positive and then 1) $x > -4$ and 2) $|\frac 7{x+4}| =\frac 7{x+4}$ and $x + 4 > 0$.
So.... $\frac 7{x+4} > 1\implies 7 < x+4 \implies 3<x $ and combining that with $x > -4$ we get.... $x > 3$.
Case 2:
If $\frac 7{x+4}$ is negative then $x+4$ is negative and 1) $x < -4$ and 2) $|\frac 7{x+4}| =\frac 7{-(x+4)}$ and $-(x + 4) > 0$.
So.... $\frac 7{-(x+4)} < 1\implies 7 < -(x+4) = -x -4\implies x< -11$ and combining that with $x< -4$ we get.... $x<-11$.
....
So Case 1: $x > 3$ and Case 2: $x < -11$.
So $x < -11$ or $x > 3$ and the solution set is:
$(-\infty, -11) \cup (3, \infty) =$
$\{$ everything that isn't between $-11$ and $3\}=$
$\mathbb R \setminus [-11,3]$.
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