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?

$$\left|\frac{x}{x+4}\right|<4$$
Squaring, you get
$$\frac{x^2}{16(x+4)^2}<1$$
Rearranging we get
$$15x^2 + 128x+256 >0$$
Using quadratic roots
$$(x+\frac{16}{5}) (x+\frac{16}{3}) >0$$
Hence  the solution is $$x \in (-\infty,-\frac{16}{3})\cup(-\frac{16}{5},\infty)$$
