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Homework Help: Inequalities Question

  1. Sep 24, 2008 #1
    Problem:
    [tex]\frac{3}{|x+1|-1}[/tex]+[tex]\frac{2}{x}[/tex]<1

    My Solution:
    There are 2 cases:
    1) x[tex]\geq0[/tex],[tex]\frac{3}{x+1-1}[/tex]+[tex]\frac{2}{x}[/tex]<1
    [tex]\frac{3}{x}[/tex]+[tex]\frac{2}{x}[/tex]<1
    And you end up with... x>5.

    2) x<0, [tex]\frac{3}{-x-1-1}[/tex]+[tex]\frac{2}{x}[/tex]<1
    [tex]\frac{3x-2x-4}{-x^{2}-2x}[/tex]<1
    ... a few reductions later...
    [tex]x^{2}[/tex]+3x-4>0
    And the solution set for this case is, -4>x>1

    My Question:
    Is there something I'm missing or something else I need to do?
     
  2. jcsd
  3. Sep 24, 2008 #2

    Mentallic

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    Homework Helper

    (edited: incorrect response)
     
    Last edited: Sep 25, 2008
  4. Sep 25, 2008 #3

    HallsofIvy

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    "-4 > x> 1" implies -4> 1! x2+ 3x- 4= (x+ 4)(x- 1)> 0 if and only if the two factors are of the same sign. x+4> 0, x- 1> 0 give x> -4 and x> 1 which are both satisfied for x> 1. x+4< 0, x-1< 0 give x< -4, x< 1 which are both satisfied for x< -4.
    x2+ 3x- 4> 0 for x< -4 OR x> 1.
     
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