<s>[INST] Solving the absolute value inequality $\big| \frac{x}{x + 4} \big| < 4$ I was given this question and asked to find $x$:
$$\left| \frac{x}{x+4} \right|<4$$
I broke this into three pieces:
$$
\left| \frac{x}{x+4} \right| = \left\{
        \begin{array}{ll}
            \frac{x}{x+4} & \quad x > 0 \\
            -\frac{x}{x+4} & \quad -4 < x < 0 \\
            \frac{x}{x+4} & \quad x < −4
        \end{array}
    \right.
$$
Solving,
$$4>\frac{x}{x+4}$$
$$4x+16>x$$
$$3x>-16$$
$$x>-\frac{16}{3}=-5.3$$
and
$$4>-\frac{x}{x+4}$$
$$4x+16>-x$$
$$5x>-16$$
$$x>-3.2$$
The answer is $x<-5.3$ and $x>-3.2$. What am I doing wrong?
 [/INST] You need to analyze each case separately, and it is always wrong to round off in mathematics without explicitly stating that it is an approximation. Do not forget that to find the solutions you need to take into account the cases! Also, you missed the case of $x = 0$.
$\def\eq{\Leftrightarrow}$
Take any real $x \ne -4$.
If $0 \le x$:
  $\left| \frac{x}{x+4} \right|<4 \eq \frac{x}{x+4} < 4 \eq x < 4x+16 \eq x > -\frac{16}{3}$.
If $-4 < x < 0$:
  $\left| \frac{x}{x+4} \right|<4 \eq -\frac{x}{x+4} < 4 \eq -x < 4x+16 \eq x > -\frac{16}{5}$.
If $x < -4$:
  $\left| \frac{x}{x+4} \right|<4 \eq \frac{x}{x+4} < 4 \eq x > 4x+16 \eq x < -\frac{16}{3}$.
Thus $\left| \frac{x}{x+4} \right|<4 \eq \left( 0 \le x \land x > -\frac{16}{3} \right) \lor \left( -4<x<0 \land x > -\frac{16}{5} \right) \lor \left( x<-4 \land x < -\frac{16}{3} \right)$.
Thus $\left| \frac{x}{x+4} \right|<4 \eq x \ge 0 \lor -\frac{16}{5}<x<0 \lor x<-\frac{16}{3}$.
["$\land$" means "and" and "$\lor$" means "or".]
</s>