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?

Your problem is very simple. :)
Why admin delete my ANS?????????????????????
Here's my Hint:
$$\left | \dfrac{x}{x+4} \right |<4 \overset{x \ne -4}{\rightarrow} -4<\dfrac{x}{x+4}<4 \Leftrightarrow \left\{\begin{matrix}
 &-4<\dfrac{x}{x+4} \\ 
 & \dfrac{x}{x+4}<4
\end{matrix}\right.
\Leftrightarrow
\left\{\begin{matrix}
 &\dfrac{5x+16}{x+4}>0 \\ 
 & \dfrac{3x+16}{x+4}>0
\end{matrix}\right.$$
$$\Leftrightarrow 
\left\{\begin{matrix}
 &x \in\left ( -\infty;-4 \right )\cup \left ( -\dfrac{16}{5};+\infty \right ) \\ 
 & x \in\left ( -\infty;-\dfrac{16}{3} \right )\cup \left ( -4;+\infty \right )
\end{matrix}\right.
\Rightarrow \boxed{x \in \left ( -\infty;-\dfrac{16}{3} \right )\cup \left ( -\dfrac{16}{5};+\infty \right).}$$
