Solve Inequality for $ |x| $ Given $$\big|\frac{(x-2)}{(x+3)}\big| < 4,$$  solve for $x.$
\ My solution
$$|x - 2| < 4|x + 3|$$
Since,
$ |x - 2| \ge |x| - |2| $ and 
$ |x + 3| \le |x| + |3| $ according to triangle inequality;
$|x| - |2| < 4|x| + 4|3| $
$-14 < 3|x|$
$|x| > \frac{-14}{3}$
Is this the final answer?

$$\left|\frac{x-2}{x+3}\right|<4$$
$$-4<\frac{x-2}{x+3}<4~~ |\cdot (x+3)^2\ne 0$$
$$-4(x+3)^2<(x-2)(x+3)<4(x+3)^2$$
$$\begin{cases}
-4(x+3)^2<(x-2)(x+3)\\
(x-2)(x+3)<4(x+3)^2
\end{cases}$$
$$\begin{cases}
0<(x+3)((x-2)+4(x+3))\\
0<(x+3)(4(x+3)-x+2)
\end{cases}$$
$$\begin{cases}
0<(x+3)(5x+10)\\
0<(x+3)(3x+14)
\end{cases}$$
$$\begin{cases}
\left[\begin{array}{}x<-3\\ x>-2\end{array}\right.\\
\left[\begin{array}{}x<-\frac{14}{3}\\ x>-3\end{array}\right.
\end{cases}$$
So the final answer is $x\in(\infty;-\frac{14}{3})\cup(-2;\infty)$
