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.

$$\frac{7}{x+4}>-1 \implies \frac{x+11}{x+4} >0$$
This means
(1): $(x+11)>0 \& x+4>0 \implies x>-4 \& x >-11 \implies x>-4$
(2): $(x+11) <0 \& x+4 <0 \implies x<-11 \& x <-4 \implies x<-11$
So the total answer is union of the two: $x\in (-\infty, -11) \cup (-4, \infty)$ or
$x<-11$ or $x >-4$.
