<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] Case $\#1:$  For $x+4\ne0$
As for real $y, a>0$
$$|y|<a\iff -a<y<a$$
$$\implies\left|\dfrac x{x+4}\right|<4\iff-4<\dfrac x{x+4}<4$$
Now $-4<\dfrac x{x+4}\iff0<\dfrac x{x+4}+4=\dfrac{5x+16}{x+4}=5\cdot\dfrac{(x+4)\left(x+\dfrac{16}5\right)}{(x+4)^2}$
$\iff(x+4)\left(x+\dfrac{16}5\right)>0$
Now if $(z-a)(z-b)>0$ with $a<b$ for real $z,$  either $z<a$ or $z>b$ (Prove this)
Similarly deal with $$\dfrac x{x+4}<4$$
Case $\#2:$ Check for $x+4=0$
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