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Inequality absolute value help

  1. Oct 4, 2013 #1
    ## x - |x-|x|| > 2 ##

    how would I go about solving something like this?

    my initial thoughts was to consider if x >= 0

    I get 2-x < 0 then x > 2 in that case

    then consider if x < 0 which I get -|x+x| > 2-x then 2x > 2-x then x > 2/3 but I'm having troubles deciding which one is correct, and if there is another way to do it (I can't seem to sketch |x-|x||

    edit: I know it's pretty obvious from looking at the equation that x > 2, but just wondering why I get the x >2/3 part from, and if I got given a question which is not obvious, then how would I know x > 2/3 would be wrong (for example), and which inequality would be the correct bit
    Last edited: Oct 4, 2013
  2. jcsd
  3. Oct 4, 2013 #2


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    You do need to consider cases, and your first bit is correct since, if [itex] x \ge 0 [/itex], the absolute value term gives
    | x - |x|| = |x - x| = 0

    and your problem reduces to [itex] x > 2 [/itex].

    For the case [itex] x < 0 [/itex], remember [itex] |x| = -x [/itex], so
    |x - |x|= |x - (-x)| = |2x| = \text{ what?}

    Simplify remembering [itex] x < 0 [/itex]: then take that simplification and plug into your original inequality statement and see what happens.
  4. Oct 4, 2013 #3
    ## |x - |x|= |x - (-x)| = |2x| = -2x ## if x < 0

    plugging this in I get x - (-2x) > 2 so x > 2/3
  5. Oct 4, 2013 #4


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    Ok, so for the inequality to be satisfied you need to have both [itex] x > \frac 2 3 [/itex] and [itex] x > 2 [/itex] true. Where on the number line do those two inequalities intersect (where do their graphs overlap)? The answer to that gives the solution set to the problem.
  6. Oct 4, 2013 #5
    x>2 obviously! So stupid

    thanks, lol.
  7. Oct 4, 2013 #6


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    "So stupid"

    Not at all. At some time we've all benefited from a little push to look at a problem with a different eye.
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