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we did this problem in class yesterday and I couldn't replicate our result at home.

|${1+x\over 1-x}$| $\le$ 1

How do I algebraically solve this and then display the solution in set notation? I tried rewriting the absolute value as

-1 $\le$ ${1+x\over 1-x}$ $\le$ 1 but then I got stuck. I can't seem to manipulate it to the point where it's obvious for what x this inequality holds true.

Any help appreciated.

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Hint

Let $x\neq 1$.

$$-1\leq \frac{1+x}{1-x}\leq 1\iff \begin{cases}-(1-x)\leq 1+x\leq 1-x&1-x>0\\ -(1-x)\geq 1+x\geq 1-x&1-x<0.\end{cases}$$

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  • $\begingroup$ This yields x>1 and x<1 so (-inf, 1) and (1, inf) but that's not true? The solution has to be (-inf, 0]. What is my misunderstanding? $\endgroup$ – Dorian James Sep 16 '17 at 11:30
  • $\begingroup$ @DorianJames: Of course it's not true ! $\endgroup$ – Surb Sep 16 '17 at 11:31
  • $\begingroup$ What am I missing? $\endgroup$ – Dorian James Sep 16 '17 at 11:35
  • $\begingroup$ @DorianJames : Solve $-(1-x)\leq 1+x\leq 1-x$ when $x<1$ and $-(1-x)\leq 1+x\leq 1-x$ when $x>1$ and conclude ! $\endgroup$ – Surb Sep 16 '17 at 11:42
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Notice that $x\ne 1$.

Get rid of the enumerator (it is positive) and you can rewrite your inequality like this $$|x+1|\leq |x-1|$$ So a distance between $1$ and $x$ is more or equal then a distance between $-1$ and $x$ so $x\leq 0$.

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