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Rational inequality

  1. Aug 23, 2014 #1
    1. The problem statement, all variables and given/known data
    Find all integer roots that satisfy (3x + 1)/(x - 4) < 1.

    3. The attempt at a solution
    I would do this:

    Make it an equation and find x such that (3x + 1)/(x - 4) = 1.

    3x + 1 = x - 4
    2x = -5
    x = -5/2

    Then check if the inequality is valid for values smaller than x and for values bigger than x.

    But this approach is not good enough as I would get [-2, +∞) {integers} as my answer.

    Any help would be really appreciated.

    I think that the answer is [-2, 3] {integers}. But could only get this with a plot.

    ---

    What should I also do so that my method is valid for "rational" inequalities?
     
  2. jcsd
  3. Aug 23, 2014 #2

    LCKurtz

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    Consider the two cases where ##x<4## and ##x>4## and work the inequalities separately.
     
  4. Aug 23, 2014 #3

    pasmith

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    [tex]
    \frac{3x + 1}{x - 4} = \frac{3(x-4) + 3(4) + 1}{x - 4} = 3 + \frac{13}{x - 4}.
    [/tex] Thus if [itex](3x + 1)/(x-4) < 1[/itex] then [itex]13/(x - 4) < - 2[/itex]. Clearly that can't be the case if [itex]x > 4[/itex] (because then [itex]13/(x - 4) > 0 > -2[/itex]) so we must have [itex]x < 4[/itex]. Is there a lower bound?
     
  5. Aug 23, 2014 #4

    ehild

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    (3x + 1)/(x - 4) < 1 can be written in the form
    [tex]\frac{(3x+1)-(x-4)}{x-4}<0[/tex]

    Simplified: [tex]\frac{2x+5}{x-4}<0[/tex]

    When is the fraction negative?

    ehild
     
  6. Aug 23, 2014 #5
    ---
    (3x + 1)/(x - 4) = 1
    3x + 1 = x - 4
    2x = -5

    x = -5/2

    ----
    (3x + 1)/(x - 4) = 1
    (3(x - 4) + 12 + 1) / (x - 4) = 1
    3 + 13/(x - 4) = 1
    13 / (x - 4) = -2

    x = 4

    ----
    Then I work with those?
    (3x + 1)/(x - 4) < 1
    Code (Text):

    - 8/3 >> false
    - 5/2 >> false
    - 7/3 >> true
        4 >> impossible
        5 >> false
     
    So the valid integers are {-2, -1, 0, 1, 2, 3}?
     
  7. Aug 23, 2014 #6

    HallsofIvy

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    An inequality can change direction where the two sides are equal or where the functions are discontinuous. Here, the first occurs where x= -5/2 and the second where x= 4. There are three intervals to be considered: x< -5/2, -5/2< x< 4, and x> 4.
    x= -3< -5/2 and (3(-3)+ 1)/(-3- 4)= (-9+ 1)/(-7)= -8/-7 is greater than 1 so NO x< -5/2 satisfies the inequality. x= 0 is between -5/2 and 4. (3(0)+ 1)(0- 4)= -1/4 is less than 1. Every x between -5/2 and 4 satisfy the inequality. x= 5 is larger than 4 and (3(5)+ 1)/(5- 4)= 15/1 is larger than 1. The integer solutions are -2, -1, 0, 1, 2, and 3.

     
  8. Aug 23, 2014 #7
    When [itex]{2x+5} < 0[/itex] and [itex]{x-4} > 0[/itex] or when [itex]{2x+5} > 0[/itex] and [itex]{x-4} < 0[/itex].

    If [itex]{2x+5} < 0[/itex], then [itex]2x<-5[/itex] and [itex]x<-\frac{5}{2}[/itex].
    and if [itex]{x-4} > 0[/itex], then [itex]x > 4[/itex]. Thus, this is impossible.

    If [itex]{2x+5} > 0[/itex], then [itex]2x>-5[/itex] and [itex]x>-\frac{5}{2}[/itex].
    and if [itex]{x-4} < 0[/itex], then [itex]x < 4[/itex]. Thus, [tex] S=\left\{x\in Z|-5 /2 < x < 4\right\}=\left\{x\in Z|-2 \le x \le 3\right\}[/tex]
     
  9. Aug 23, 2014 #8
    Big thanks to all of you, with special mention to HallsOfIvy for answering my question.

    Q.: "What should I also do so that my method is valid for "rational" inequalities?"
    A.: An inequality can change direction where the two sides are equal or where the functions are discontinuous.
     
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