$|x| + |x-1| = 3$ how come its cases? $$|x| + |x-1| = 3$$
in my textbook, they say that for this equation, there are 3 cases:  
$x\geq1$,
$0 \leq x < 1$ and
$ x < 0$
where do these come from and why?  i thought, there are 4 cases
$x$ positive
$x-1$ positive  
$x$ is negative
$x-1$ is negative  
$x$ positive
$x-1$ is negative
$x$ is negative
$x-1$ is positive  
 A: In the first case, both $x$ and $x-1$ are non-negative. In the second case, $x$ is non-negative but $x-1$ is negative.  In the last case, both $x$ and $x-1$ are negative.
A: The problem is the absolute value. Sometimes $|a|$ is $a$ and sometimes $|a|$ is $-a$. 
Here we have the cases


*

*$x\ge1$, where $|x-1|=x-1$ and collatrally also $|x|=x$.

*$x<0$, where $|x|=-x$ and also $|x-1|=1-x$

*the intermediate case $0\le x<1$, where $|x|=x$, but $|x-1|=1-x$.


In each case, replacing the absolute value expression with the corresponing absoluteness-less expression for the case in question gives us a simple equation that can be solved.
A: The 4 cases are entirely correct, but with a little bit of reasoning, they can be reduced to the 3 described by your book:
Let's see how the first modulus (A) behaves:


*

*$|x| = x$, for $x >= 0$

*$|x| = -x$, for $x < 0$


Now for the second (B):


*

*$|x-1| = x-1$, for $x - 1 >= 0$ ($x >= 1$)

*$|x-1| = 1-x$, for $x - 1 < 0$ ($x < 1$)


The equation becomes:


*

*A1B1: $x + x-1 = 3$, for $x >= 0$ and $x >= 1$

*A1B2: $x + 1-x = 3$, for $x >= 0$ and $x < 1$

*A2B1: $-x + x-1 = 3$, for $x < 0$ and $x >= 1$

*A2B2: $-x + 1-x = 3$, for $x < 0$ and $x < 1$


You can clearly see that A2B1 is invalid because $x$ can't be smaller than $0$ and larger than $1$ at the same time (which is the case you have found but has been silently excluded from the book).
A1B1 and A2B2 conditions can also be simplified to $x >= 1$ / $x < 0$ respectively, coming up with the cases from the book.

Another way to think about your problem is to see what intervals to analyze for $x$ by seeing the interesting values. $-Infinity$, $Infinity$ are always interesting and the other two interesting values are $0$ and $1$ (the values where the modulus functions change behavior).
Plotting the Numbers axis with these values highlighted looks like this:
$-Infinity$ ________ 0 ____ 1 _______ $Infinity$
and this immediately makes it obvious that you have 3 intervals to check.
A: The absolute value function does different things depending on whether the quantity inside the bars is positive or negative.  In this case:


*

*$x-1$ is positive if $x > 1$, in which case $|x-1|$ is just equal to $x-1$.

*$x-1$ is negative if $x<1$, in which case $|x-1$ is equal to $-(x-1)$ (or if you prefer $1-x$).

*On the other hand $|x|$ equals just $x$ if $x>0$.

*And $|x|$ equals $-x$ if $x<0$.


If you look at those four statements you'll notice that there are really three regions in which the behavior of the expression needs to be considered separately.
Hope this helps.
A: |x| is either +x or -x, depending on the value of x. Same with |x-1|. So for each absolute value, you consider the case that the argument is >= 0, and the case that the argument is <= 0. That's two cases for each absolute value, and two altogether. 
If x >= 1, then both x and x-1 are >= 0, and |x| = x, |x-1| = x-1. 
If 0 <= x < 1, then x >= 0 but x-1 < 0, so |x| = x, |x-1| = - (x-1) 
and so on. There is no possible value x where x < 0 but x-1 >= 0, so only three of the four cases are possible. 
