# Inequality of absolute values

• B
blueblast

## Main Question or Discussion Point

Hi there,

I'm having trouble understanding this math problem:

|x| + |x-2| = 2

I understand you need different "cases" in order to solve this. For example, cases for when x is less than 0, when x-2 is less than 0, etc.

Thanks,

blueblast

Doc Al
Mentor
For each of those cases, write an equation in terms of x (without the absolute value signs) and try to solve it. For example, when x < 0, how would you write |x| ?

Last edited:
blueblast
Yes, so far, this is what I got:

There are a total of four cases:

1. x + x-2 = 2
2. x + -(x-2) = 2
3. -x + x-2 = 2
4. -x + -(x-2) = 2

Simplified, this is:

1. x = 2
2. 2 = 2 (all reals)
3. -2 = 2 (no solution)
4. x = 0

Not sure where to go from there.

S.G. Janssens
It could help to make a drawing of the real line. On that line, indicate the points ##0## and ##2##.

Now note that ##|x|## equals the distance from ##x## to ##0## and, likewise, ##|x - 2|## equals the distance from ##x## to ##2##. Apparently the sum of these two distances should equal ##2##, for ##x## to be a solution.

So, could any ##x < 0## be a solution of your equation? Could any ##x > 2## be a solution?
You have now narrowed down the possible solutions to the interval ##[0,2]##. For ##x \in [0,2]## your equation becomes
$$|x| + |x - 2| = x + (2 - x) = 2$$
For which ##x \in [0,2]## is this equation satisfied?

Doc Al
Mentor
There are a total of four cases:
You'll need to define those cases. (Think in terms of ranges.) For example, for x < 0, your equation can be written as:
-x + 2 - x = 2
For which the solution is x = 0, which violates x < 0. Thus we can exclude the range x < 0.

And so on....