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Solving (x+1)/(2x-3) > 2; why do we take 2x-3>0 while obtaining the solution?

  1. Jan 18, 2012 #1
    I mean it's okay to take 2x-3≠0 as a necessary condition, but I can't actually grasp the fact that we have to take the denominator to be '>0' and not merely '≠0'.
    A little guidance will be very much appreciated :)
     
  2. jcsd
  3. Jan 18, 2012 #2

    Stephen Tashi

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    Science Advisor

    If you solve the inequality in a deductive manner, you don't only look at the case 2x -3 > 0. You look at all the possible cases for the sign of the numerator and denominator of the fraction (x+1)/(2x+3). You say that the cases when one is positive and the other is negative can't povide any solutions since the left hand side would be negative, so less than 2. In another case, both the numerator and denominator are positive. In that case x+1 > 0 and 2x+3 > 0, so x> -1 and 2x > -3, x > -3/2.

    You have to look at all the cases to work the problem in a systematic manner. Perhaps the materials you are looking at know some clever shortcut.
     
  4. Jan 20, 2012 #3
    The above is not one of the correct approaches.


    Mine:


    Subract 2 from each side, and that gives a new fraction that is greater
    than 0. Then find the critical numbers to establish the candidate
    intervals from which to choose for the solution:


    After subtracting 2 and simplifying the fraction:


    [itex]\dfrac{-3x + 7}{2x - 3} > 0[/itex]


    Critical numbers:

    -3x + 7 = 0

    3x = 7

    x = 7/3

    -------------------

    2x - 3 = 0

    2x = 3

    x = 3/2

    -------------------


    Then the candidate intervals to choose from for the solution are:

    (-oo, 3/2), (3/2, 7/3), and (7/3, oo).


    Upon checking test values, it is shown that the solution is:


    (3/2, 7/3),


    or equivalently, as


    3/2 < x < 7/3
     
  5. Jan 20, 2012 #4

    Deveno

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    Science Advisor

    since the fraction on the left is greater than 2 > 0, both the numerator and the denominator must have the same sign.

    let's look at the case where both the numerator and the denominator are < 0:

    x+1 < 0 => x < -1.
    2x - 3 < 0 => x < 3/2.

    so for both these to be true, we take the more restrictive requirement: x < -1.

    since with that assumption, we have 2x - 3 < 0 as well, we have:

    x + 1 < 4x - 6 (when we multiply by a negative number, we reverse the inequality)
    1 < 3x - 6
    7 < 3x
    7/3 < x

    but this is a contradiction, x can't be BOTH < -1 AND > 7/3.

    so the case where both numerator and denominator are negative, does not lead to any valid solutions.

    so we may assume, without loss of generality (in THIS particular case) that both numerator and denominator are > 0.
     
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