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Solving absolute value inequalities

  1. May 2, 2012 #1
    1. The problem statement, all variables and given/known data
    [abs(x)]/[abs(x+2)]<2


    2. Relevant equations



    3. The attempt at a solution

    case 1: [abs(x)]/[abs(x+2)]<2
    case 2: [abs(x)]<2[abs(x+2)]
    is this right so far?
    if so, why is there two cases
    and what do i do next?
     
  2. jcsd
  3. May 2, 2012 #2

    HallsofIvy

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    What you have written is NOT two cases. It is two versions of the same inequality.

    Because an absolute value is always positive, multiplying both sides of "case 1" by |x+2| does not change the direction of the inequality sign and leads to "case 2".

    To solve this inequality, you should consider three cases:
    a) [itex]x\le -2[/itex] so that x and x+2 are both less than 0. |x+ 2|= -(x+2) and |x|= -x.
    b) [itex]-2< x\le 0[/itex] so that x+ 2 is positive but x is still less than 0. |x+ 2|= x+ 2 and |x|= -x.
    c) [itex]0< x[/itex] so that both x and x+ 2 are both positive. |x+ 2|= x+ 2 and |x|= x.
     
  4. May 2, 2012 #3

    SammyS

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    Those are not two different cases.

    The inequality [itex]\displaystyle\frac{|x|}{|x+2|}<2[/itex] is equivalent to [itex]\displaystyle|x|<2|x+2|\ .[/itex]
     
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