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


Hint:
The domain of validity of this inequation is $\mathbf R\smallsetminus\{-1\}$.
On this domain,
\begin{align} \biggl|\frac {x-1}{x+1}\biggr|<2  &\iff\biggl(\frac {x-1}{x+1}\biggr)^{\!2}<2^2
\iff \bigl(2(x+1)\bigr)^2-(x-1)^2>0\\[1ex]
&\iff (3x+1)(x+3)>0.
\end{align}
