Help to solve absolute value inequality

Question

The inequality I have is $\frac {\mid x-1 \mid} {(x+2)} <1 $ what I'm not sure is how I am supposed to proceed. I cannot multiply by (x+2) because it is unknown whether it is positive or negative. What I have done is changed it to $\frac {1}{\mid x-1 \mid} > \frac {1}{x+2}$ but I don't know if now I can solve it as $\frac {1}{\mid x-1 \mid} > \frac {1}{x+2}$ and $\frac {1}{\mid x-1 \mid} < - \frac {1}{x+2}$. I'm just a bit confused as to when I should be solving inequalities by the theorem of -ax and x<-a or whether to graph it and solve by each case, positive or negative.

Answer 1

Split the problem into two cases: $x-1<0$ and $x-1\ge0$.

First case, $x<1$

$$ \begin{cases} \dfrac{1-x}{x+2}-1<0 \\[4px] x<1 \end{cases} $$ The top inequality becomes $$ \frac{2x+1}{x+2}>0 $$ that's satisfied for $x<-2$ or $x>-1/2$. Together with $x<1$, you get $$ \boxed{x<-2\quad\text{or}\quad -\frac{1}{2}<x<1} $$

Second case, $x\ge1$

$$ \begin{cases} \dfrac{x-1}{x+2}-1<0 \\[4px] x\ge1 \end{cases} $$ The top inequality becomes $$ \frac{3}{x+2}>0 $$ that's satisfied for $x>-2$. Together with $x\ge1$, you get $$ \boxed{x\ge1} $$

Conclusion

The inequality is satisfied for $$ \boxed{x<-2\quad\text{or}\quad x>-\frac{1}{2}} $$

Answer 2

Hint:
You can in these situations break down the problem by using two cases: One in which $x+2\gt0$, and one in which $x+2\lt 0$.