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Inequalities involving division of two absolute values

  1. Sep 12, 2011 #1
    1. The problem statement, all variables and given/known data

    How to solve x for these inequality?

    2. Relevant equations

    |x-2|/|x+3|> (x+2) / (x+1)

    3. The attempt at a solution

    (x - 2)/(x + 3) > (x + 2) / ( x+1)

    the left side holds the condition that is x >= 2

    however, I wonder the next step. should I crossly multiply so I get
    [(x - 2)((x + 1) / (x+3)] > (x + 2)(x +3) /(x+1) and find the value of x?

    or there is another condition?
  2. jcsd
  3. Sep 12, 2011 #2

    Stephen Tashi

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    Science Advisor

    Whatever you did, isn't what we call "cross multiplication".

    There are many other combinations of conditions. For rexample you multiply both sides of an inequality by (x-1) you must consider the cases:
    Case 1) x -1 > 0
    Case 2) x-1= 0
    Case 3) x-1 < 0. (If you multiply both sides of an inequality by by a negative number, you must reverse in inequality sign)

    When you begin the problem you should enumerate the cases that must be considerd for the absolute values signs:

    Case 1) [itex] x-2 \ge 0 ; x + 3 \ge 0 [/itex] implies |x-2] = x-2 and |x+3] = x + 3
    Case 2) [itex] x-2 \ge 0; x + 3 < 0 [/itex] is impossible
    Case 3) [itex] x -2 < 0; x + 3 \ge 0 [/itex] implies |x-2| = -(x-2) and |x+3|= x + 3
    Case 4) [itex] x-2 < 0; x + 3 < 0 [/itex] implies |x-2| = -(x-2) and |x+3| = -(x+3)

    It's a pain in the neck to solve problems like this deductively. is that what your course materials want you to do? Or do they want you to solve this by graphing it?
  4. Sep 13, 2011 #3


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    Homework Helper

    To solve an inequality with absolute values, first investigate all possible cases. The sign of a factor is critical if its absolute value appears in the inequality or if it is in a denominator.. Also exclude the cases when a denominator is zero.
    Here you have four intervals :x<-3, -3<x<-1,-1<x<2, and x≥2.

    If x<-3: x+3<0, x-2<0, x+1<0, x+2<0,
    if -3<x<-1: x+3>0, x+1<0, ....

    Eliminate the absolute values in each interval, according to the signs.
    Then you can cross-multiplying, but take care: the inequality turns to opposite if you multiply with a negative quantity.


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