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 $x\in(\infty;-\frac{14}{3})\cup(-2;\infty)$
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 $$ \left|\frac{x-2}{x+3} \right| < 4 \text{ for } -\infty < x < -14/3, -2 < x < \infty.$$
