Find the set of values of x for which $\lvert \frac {x-1}{x+1} \rvert <2 $ How to solve this inequality question involving modulus?
I can’t get the same answer as the book [answer below]
I know the properties of absolute values, that is
If $\lvert x \rvert <k$ , then $-k < x < k$.
So for this question, this is my working:
$-2<\lvert \frac {x-1}{x+1} \rvert <2 $
When $-2<\lvert \frac {x-1}{x+1} \rvert $
$-2x-2 < x-1$
$-1<3x$
$x>- \frac {1}{3}$
And when $\lvert \frac {x-1}{x+1} \rvert <2 $
$x-1<2x+2$
$x>-3$
So, I got $x> - \frac {1}{3}\:$ or $\:x>-3$
However,  The **answer given is  $\{ {x\mid x<-3\: \text{ or }\: x> -\frac {1}{3}, ∈ℝ} \}$  **
I don't think the answer from the book is wrong, since the following graph confirms that the book answer is correct.
Please show me how to get the answer.
Thank you


The problem with the analysis of the second part is that you have taken $|\frac{x-1}{x+1}| 
 = \frac{x-1}{x+1}$ and done the analysis. This is false : what about those $x$ for which we have $|\frac{x-1}{x+1}| = \boxed{-\frac{x-1}{x+1}}$?

You need to break your analysis according to where $|\frac{x-1}{x+1}| = \frac{x-1}{x+1}$ and where it is the negative of the expression. After this, you can work separately on each component.

For example, when is $\frac{x-1}{x+1} > 0$? It happens if and only if both numerator and denominator are positive, or both are negative. One checks that this comes to $x<-1$ or $x>1$.
In these intervals, one solves $-2<\frac{x-1}{x+1}<2$. (i.e. one solves this, then takes the intersection with $x<-1 \cup x>1$).
On the interval $-1 < x \leq 1$, one solves $-2 < -\frac{x-1}{x+1}< 2$. Then we can put them together to finish.
In short, your argument falls because you assumed that the value $\frac{x-1}{x+1}$ was positive all the time : instead you must break it into where it is negative, where it is positive and then work separately on both parts.
