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

 A: 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.
A: 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}
A: If $x=a+ib$ where $a,b$ are real
We have $$\sqrt{(a-1)^2+b^2}<2\sqrt{(a+1)^2+b^2}$$
Squaring and simplifying we get 
$$b^2+a^2+\dfrac{10a}3+1>0$$
$$b^2+\left(a+\dfrac53\right)^2>\dfrac{25}9-1=\left(\dfrac43\right)^2$$
So, $x$ needs to lie outside the circle $$y^2+\left(x+\dfrac53\right)^2=\left(\dfrac43\right)^2$$
If $b=0,x=a$ is real
$$\left(a+\dfrac53\right)^2>\left(\dfrac43\right)^2\iff\left(a+\dfrac13\right)(a+3)>0$$
which happens 
if $a>$max$\left(-\dfrac13,-3\right)$ 
or if $a<$min$\left(-\dfrac13,-3\right)$ 
In any case, $a\ne-1$
