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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.

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    $\begingroup$ Well if $x+2$ is negative, anyway the LHS is lesser and the inequality holds. So you may assume it isn't and multiply to check for any other solution... $\endgroup$ – Macavity Apr 22 '16 at 19:38
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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 $$ \boxed{x<-2\quad\text{or}\quad x>-\frac{1}{2}} $$

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

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  • $\begingroup$ I tried that and got the solutions $ (\-infty,-2)U (1,\infty)$ and $(-\infty,-2) U (-1/2,1)$ for $x>-2$ and $x<-2$ respectively. Should I now take the intersection of these domains or the union? $\endgroup$ – Voltai Apr 22 '16 at 19:46
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    $\begingroup$ If you assume $ x<-2$ and get some domain $R$, then you must only keep $R\cap (-\infty,-2)$ for that case. Also, at first glance you should check your solution for the case $x<-2$ anyway. $\endgroup$ – Clement C. Apr 22 '16 at 19:48
  • $\begingroup$ What mistake did I make for $x<-2$ ? $\endgroup$ – Voltai Apr 22 '16 at 20:08
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    $\begingroup$ If $x < -2$, $\frac{\lvert x-1 \rvert}{x+2} = \frac{1-x}{x+2}$. if you solve the inequality then, you get $x < -\frac{1}{2}$ (not $x\in(-1/2,1)$), so that basically the solution (since you assumed $ x < -2$ in the first place) is $ x < -2$. $\endgroup$ – Clement C. Apr 22 '16 at 20:25

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