Solve $|1 + x| < 1$ I'm trying to solve $|1 + x| < 1$. The answer should be $ -2 < x < 0$ which wolframalpha.com agrees with.
My approach is to devide the equation to:
$1+x < 1$ and $1-x < 1$ and then solve those two:
$ 1+x < 1 $
$ x < 0 $
$ 1 - x < 1$
$ -x < 0$
$x > 0$
And this gives me $ x = 0$. Where am I going wrong?

Note that:
$$
|a|=
\begin{cases}
\;\;\;a &\text{if} \quad a\geq 0,\\
-a &\text{if} \quad a< 0.
\end{cases}
$$
Hence,
$$
|x+1|=
\begin{cases}
\;\;\;x+1 &\text{if} \quad x\geq -1,\\
-x-1 &\text{if} \quad x< -1.
\end{cases}
$$


*

*If $x+1\geq 0$ (or $x\geq -1$) then $|1+x|<1$ is equivalent to $1+x<1$ or $x<0$. Then in this case $-1\leq x<0$.

*If $x+1<0$ (or $x< -1$) then $|1+x|<1$ is equivalent to $-1-x<1$ or $x>-2$. Then in this case $-2<x<-1$.
In both cases we have $-2<x<0$. 
