Solve $|1 + x| < 1$

Question

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?

Answer 1

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 $-2<x<0$.

Answer 2

You have $1-x<1$, which is incorrect. It should be $1+x>-1$ instead. This gives us $x>-2$.

Answer 3

Every exercise dealing with $| \cdot |$ can be solved as follows:

\begin{align}|1+&x| < 1 \\ (\iff) -1 < 1&+x < 1 \\ (\iff) -2 < &x < 0 \end{align}

Answer 4

$|x-(-1)|<1$ says the distance between $x$ and $-1$ needs to be less than $1$. So $x$ is in $(-2,0)$. With practice, you can do these kinds of problems without step-by-step algebra.