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Homework Help: Solving absolute value inequalities

  1. May 2, 2012 #1
    1. The problem statement, all variables and given/known data

    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


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    Science Advisor

    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


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    Staff Emeritus
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    Homework Helper
    Gold Member

    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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