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Help On Absolute Values

  1. Sep 23, 2005 #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.
     
  2. jcsd
  3. Sep 23, 2005 #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?
     
  4. Sep 23, 2005 #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: Sep 23, 2005
  5. Sep 23, 2005 #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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