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Homework Help: Solving an absolute value inequality

  1. Apr 4, 2013 #1
    1. The problem statement, all variables and given/known data

    lx/(x-2)l < 5

    2. Relevant equations

    3. The attempt at a solution
    x/(x-2) < 5
    x< 5x-10
    10 < 4x
    5/2 < x

    x/(x-2) > -5
    x > -5x+10
    6x > 10
    x > 5/3

    The answer is x < 5/3 and x > 5/2
    so where did I go wrong on the second one?
  2. jcsd
  3. Apr 4, 2013 #2


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    For P a positive real number, [itex]\ \displaystyle \left| f(x) \right|<P\ [/itex] is equivalent to [itex]-P<f(x)<P\ .[/itex]

    More to the point, [itex]\displaystyle \ \frac{x}{x-2}<5\ [/itex] is not equivalent to [itex]\ x<5(x-2)\ .\ [/itex] In fact, if x-2 is negative, then you need to change the sense of the inequality .

    You're much better off to write [itex]\displaystyle \ \frac{x}{x-2}<5\ [/itex] as [itex]\displaystyle \ \frac{x}{x-2}-5<0\,,\ [/itex] then write the left-hand side as a single rational expression by using a common denominator .
  4. Apr 4, 2013 #3
    Okay so I simplified that farther to achieve this, 2(3x-5) / x-2 > 0. So now how do I i know if X> 5/3 OR X<5/3, also wouldn't there be an inequality in the denominator, such as x>2 or x<2 . Also how do display the mathematical operation like that on the forum?
  5. Apr 4, 2013 #4


    Staff: Mentor

    The answer can't possibly be "x < 5/3 and x > 5/2". There aren't any numbers that are simultaneously smaller than 5/3, and larger than 5/2.

    How did you get this? I'm assuming you are starting from x/(x - 2) - 5 < 0, and took Sammy's advice to combine the two terms on the left into a single rational expression.

    Also, if 2(3x-5) / x-2 is supposed to mean this:
    $$ \frac{2(3x - 5)}{x - 2}$$
    then you need to write it with parentheses around the entire denominator, like so:
    2(3x-5) / (x-2)
    Otherwise, we would interpret this to mean this:
    $$ \frac{2(3x - 5)}{x} - 2$$
  6. Apr 4, 2013 #5


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    So you have one expression, 2(2x-5), divided by another, (x-2), and the result is positive. How does the sign of the numerator compare to the sign of the denominator ?
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