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Abs((x-1) / (x+2)) < 1

  1. Oct 18, 2010 #1
    I've tried squaring both sides, then moving the RHS to the LHS, then factorizing according to a2-b2=(a+b)(a-b).

    I simplified, and got
    ((2x+1)/(x+2))((-3/(x+2)) < 0

    Then there are 2 cases:
    ((2x+1)/(x+2)) < 0 and ((-3/(x+2)) > 0

    or

    ((2x+1)/(x+2)) > 0 and ((-3/(x+2)) < 0

    I'm not really sure how to go from here...
    Please reply ASAP!

    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 18, 2010 #2
    No need to square.

    Don't forget: [itex]|x| < y[/tex], means [itex]\pm x < y[/tex]
     
  4. Oct 18, 2010 #3
    thanks for telling me! now i see how much easier it is :)
     
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