Solving |x-1|+|x-2|>1 with absolute values

[JasonRox]

This is easy, but for some reason I can't grasp the idea.

$|x-1|+|x-2|>1$

I know it means that the distance between x and 2, and x and 1 is larger than 1.

It isn't school related, and I did search online, but they are simple ones like...

$|x-1|>1$

Can anyone help me?

Thanks.

 

[Päällikkö]

I'll use the simpler one as an example:
$$|x-1| = \left\{ \begin{array}{l}<br /> x-1, \mbox{when } x \ge 1 \\<br /> -(x-1), \mbox{when } x < 1 \\<br /> \end{array} \right.$$

The equation in parts:
$$x \ge 1:$$
$$x-1>1$$
$$x>2$$

$$x < 1:$$
$$-(x-1)>1$$
$$-x+1>1$$
$$x<0$$

Combining the answers we get:
$$x<0 \vee x>2$$

This said, can you solve the more complex one?

 

[arildno]

Construct 3 systems of inequalities the union of whose solutions will be the solutions of your original inequality:

1. $$|x-1|+|x-2|>1, x>{2}$$
Using the info from the second inequality to simplify the first yields:
x-1+x-2>2\to{x}>2
This system of inequalities is fulfilled for $$x>{2}$$

2. $$|x-1|+|x-2|>1, 1\leq{x}\leq{2}$$
Using the second inequality to simplify the first:
$$x-1+2-x>1\to{1}>1$$
That is, no solutions exist.

3. $$|x-1|+|x-2|>1,x\leq{1}$$
Using info from the second inequality to simplify the first, we get:
$$3-2x>1$$ that is, x<1, along with the inequality $$x<{1}$$
This means that x<1 is the solutions in this subdomain.

Thus, x<1 or x>2 are solutions to your original inequality.

 

[JasonRox]

So, basically if I get something along the lines of...

|x-a|+|x-b|>c

I should just have 3 cases.

Thanks, arildno.

I might be back with some more.