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Help On Absolute Values

  • Thread starter JasonRox
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  • #1
JasonRox
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This is easy, but for some reason I can't grasp the idea.

[itex]|x-1|+|x-2|>1[/itex]

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...

[itex]|x-1|>1[/itex]

Can anyone help me?

Thanks.
 

Answers and Replies

  • #2
Päällikkö
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I'll use the simpler one as an example:
[tex]|x-1| = \left\{ \begin{array}{l}
x-1, \mbox{when } x \ge 1 \\
-(x-1), \mbox{when } x < 1 \\
\end{array} \right.[/tex]

The equation in parts:
[tex]x \ge 1:[/tex]
[tex]x-1>1[/tex]
[tex]x>2[/tex]

[tex]x < 1:[/tex]
[tex]-(x-1)>1[/tex]
[tex]-x+1>1[/tex]
[tex]x<0[/tex]

Combining the answers we get:
[tex]x<0 \vee x>2[/tex]

This said, can you solve the more complex one?
 
  • #3
arildno
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Construct 3 systems of inequalities the union of whose solutions will be the solutions of your original inequality:

1. [tex]|x-1|+|x-2|>1, x>{2}[/tex]
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 [tex]x>{2}[/tex]

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

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

Thus, x<1 or x>2 are solutions to your original inequality.
 
Last edited:
  • #4
JasonRox
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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.
 

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