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?

from the graph  $$f(x) = \frac{x-2}{x+3} = 1-\frac5{x+3}$$ you can see that it has vertical asymptote $x = 3$ and a horizontal one $y = 1.$  the function $f$ is decreasing on $\-infty, -3)$ and increasing on $-3, \infty).$
solving $f(x) = 4$  gives $x = -14/3$ and $f(x) = -4$ gives $x = -2.$
therefore $$ \left|\frac{x-2}{x+3} \right| < 4 \text{ for } -\infty < x < -14/3, -2 < x < \infty.$$ 
