$|x| + |x-1| = 3$ how come its cases?

Question

$$|x| + |x-1| = 3$$

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

Answer 1

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.

Answer 2

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.

Answer 3

The absolute value function does different things depending on whether the quantity inside the bars is positive or negative. In this case:

1. $x-1$ is positive if $x > 1$, in which case $|x-1|$ is just equal to $x-1$.
2. $x-1$ is negative if $x<1$, in which case $|x-1$ is equal to $-(x-1)$ (or if you prefer $1-x$).
3. On the other hand $|x|$ equals just $x$ if $x>0$.
4. 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.

Answer 4

|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.

Answer 5

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:

1. $|x| = x$, for $x >= 0$
2. $|x| = -x$, for $x < 0$

Now for the second (B):

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

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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.