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?
 A: Your mistake was not considering the restrictions on $x$ when you solved the individual cases.
Case 1:  $x \geq 0$
If $x > 0$, then $x + 4 > 0$, so the direction of the inequality is preserved if we multiply by $x + 4$.  Hence,
\begin{align*}
\frac{x}{x + 4} & <  4\\
x & < 4(x + 4)\\
x & < 4x + 16\\
-16 & < 3x\\ 
-\frac{16}{3} & < x
\end{align*}
as you found.  However, the restriction that $x \geq 0$ means $x \geq 0$ and $x > -\dfrac{16}{3} \Longrightarrow x \geq 0$.  You did not account for the restriction $x \geq 0$.
Also, note that $-\dfrac{16}{3} \approx 5.3$.  The statement you made that $-\dfrac{16}{3} = 5.3$ is false since the decimal approximation you used does not actually equal the fraction.
Case 2:  $-4 < x < 0$
Since $-4 < x < 0$, $x + 4 > 0$.  Thus, the direction of the inequality is preserved if we multiply both sides of the inequality by $x + 4$.  Hence,
\begin{align*}
-\frac{x}{x + 4} & < 4\\
-x & < 4(x + 4)\\
-x & < 4x + 16\\
-16 & < 5x\\
-\frac{16}{5} & < x
\end{align*}
as you found.  The restriction that $-4 < x < 0$ means $-4 < x < 0$ and $-\dfrac{16}{5} < x \Longrightarrow -\dfrac{16}{5} < x < 0$.  You did not account for the restriction $x < 0$.
Case 3:  $x < -4$
If $x < -4$, then $x + 4 < 0$, so the direction of the inequality is reversed if we multiply both sides of the inequality by $x + 4$, which you did not take into account.
\begin{align*}
\frac{x}{x + 4} & < 4\\
x & > 4(x + 4)\\
x & > 4x + 16\\
-16 & > 3x\\
-\frac{16}{3} & > x
\end{align*}
The restriction that $x < -4$ means that $x < -4$ and $x < -\dfrac{16}{3} \Longrightarrow x < -\dfrac{16}{3}$.  
Combining the results of the three cases yields 
$$x \geq 0~\text{or}-\frac{16}{5} < x < 0~\text{or}~x < -\frac{16}{3} \Longrightarrow x > -\frac{16}{5}~\text{or}~x < -\frac{16}{3}$$
Hence, the solution set is 
$$S = \left(-\frac{16}{5}, \infty\right) \cup \left(-\infty, -\frac{16}{3}\right)$$
A: $$\left|\frac{x}{x+4}\right|<4$$
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 $$x \in (-\infty,-\frac{16}{3})\cup(-\frac{16}{5},\infty)$$
A: 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".]
A: 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 $x+4=0$
A: Your problem is very simple. :)
Why admin delete my ANS?????????????????????
Here's my Hint:
$$\left | \dfrac{x}{x+4} \right |<4 \overset{x \ne -4}{\rightarrow} -4<\dfrac{x}{x+4}<4 \Leftrightarrow \left\{\begin{matrix}
 &-4<\dfrac{x}{x+4} \\ 
 & \dfrac{x}{x+4}<4
\end{matrix}\right.
\Leftrightarrow
\left\{\begin{matrix}
 &\dfrac{5x+16}{x+4}>0 \\ 
 & \dfrac{3x+16}{x+4}>0
\end{matrix}\right.$$
$$\Leftrightarrow 
\left\{\begin{matrix}
 &x \in\left ( -\infty;-4 \right )\cup \left ( -\dfrac{16}{5};+\infty \right ) \\ 
 & x \in\left ( -\infty;-\dfrac{16}{3} \right )\cup \left ( -4;+\infty \right )
\end{matrix}\right.
\Rightarrow \boxed{x \in \left ( -\infty;-\dfrac{16}{3} \right )\cup \left ( -\dfrac{16}{5};+\infty \right).}$$
A: Divide the interval into 3 pieces.
X is between (-4,0)
X is greater than 0
X is less than -4
On each interval try to figure out whether the sign is positive or negative.
$$\left (\frac{|x|}{|x+4|}\right)$$
