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.

As noticed you went wrong multiplying for a quantity which can also be negative.
Using that for $A,B>0$ then
$$|A|<B \iff \left|\frac1{A}\right|>\frac1B $$
a very effective way to proceed in this case is the following
$$\left|\frac{7}{x+4}\right|<1 \iff \left|\frac{x+4}{7}\right|>1 \iff |x+4|>7$$
$$\iff x+4<-7 \quad \lor \quad x+4 >7 $$
$$\iff x<-11 \quad \lor \quad x>3 $$
We can also proceed by your way but we need to consider two cases, that is

*

*For $x+4>0 \iff x>-4$
$$\left|\frac{7}{x+4}\right|<1\iff \frac{7}{x+4}<1\iff x+4>7 \iff x>3$$

*

*For $x+4<0 \iff x<-4$
$$\left|\frac{7}{x+4}\right|<1\iff \frac{7}{-(x+4)}<1\iff -(x+4)>7\iff x+4<-7 \iff x<-11$$
