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Inequality with two absolute values

  1. Oct 28, 2012 #1
    1. The problem statement, all variables and given/known data
    Find all real values of x that satisfy the following inequality.

    2. Relevant equations
    [itex]|x-3| > |x + 1|[/itex]



    3. The attempt at a solution
    Splitting up the inequality in to cases I get:

    1. [itex] |x-3| > x + 1[/itex] and 2. [itex]|x-3| < -x - 1[/itex]


    1. [itex] x-3 > x + 1[/itex] or [itex]x-3 < -x - 1[/itex]

    2. [itex] x-3 < x + 1[/itex] or [itex]x-3 > -x - 1[/itex]

    The solutions to these inequalities just don't make sense. I've done something wrong with splitting them up, but I'm not sure what.
     
  2. jcsd
  3. Oct 28, 2012 #2

    MarneMath

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    Well! That looks fine to me. I understand that one case gives you 'no real information' but the other one should tell you something about x = 1. my advice plot these two graphs and notice their relation. Secondly you can notice that one is a V shape and the other is an upside down V, and infer that they'll only intersect at 1 point.
     
  4. Oct 28, 2012 #3

    SammyS

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    You should not have flipped the inequality sign around for item 2 above.

    |x + 1| = x + 1 when x+1≥0, that is to say, when x ≥ -1 .

    Similarly, |x + 1| = -x - 1 when x+1≤0, that is to say, when x ≤ -1 .

     
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