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Help with inecuation

  1. Sep 17, 2012 #1
    Hello!

    a)[itex]\frac{1}{2}<\frac{5}{3x+42}<\frac{3}{2}[/itex]

    b)[itex]\frac{1}{2}<\frac{3x+1}{3x+4}<\frac{3}{2}[/itex]

    c)[itex]\frac{1}{2}<\frac{3x+1}{3x+4}<\frac{2}{3}[/itex]

    Can you tell me how I can resolve it?

    In c), x must be less than 1/2 for the inequality is fulfilled, so I now x<1/2,

    but when I pass 2x-1 to multipliy with 1 "(3<1·2x-1)" Why I have not change "<" by ">" The denominator is negative, when is negative I thought I had change it.

    The solution set is [itex](-\infty,\frac{1}{2})\cup(2,+\infty)[/itex]
    for x> 2 not to have to change the sign to move the denominator to multiply by 1.

    Thank you very much for your help :)
     
    Last edited: Sep 17, 2012
  2. jcsd
  3. Sep 17, 2012 #2
    If x<-4/3 then (3x+1)/(3x+4) is always greater than one. When x=0, (3x+1)/(3x+4)=1/4. When x=infinity, (3x+1)/(3x+4)=1. There's some region, while x is positive, that the expression is within your region of interest.
     
  4. Sep 17, 2012 #3

    SammyS

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    Re: help with inecuation (inequality)

    What method have you been taught to solve rational inequalities?

    I like to solve for equality & then use the fact that the rational function is continuous, except where the denominator is zero. Solving for equality eliminates the need to separate into cases depending upon the sign of the denominator.

    Look at (c): [itex]\displaystyle \frac{1}{2}<\frac{3x+1}{3x+4}<\frac{2}{3}[/itex]

    Solve: [itex]\displaystyle \frac{1}{2}=\frac{3x+1}{3x+4}[/itex]

    [itex]3x+4=6x+2[/itex]

    [itex]\displaystyle x=\frac{2}{3}[/itex]

    Solve: [itex]\displaystyle \frac{3x+1}{3x+4}=\frac{2}{3}[/itex]

    [itex]9x+3=6x+8[/itex]

    [itex]\displaystyle x=\frac{5}{3}[/itex]​

    Check easy to use numbers: one to the left of 2/3 one between 2/3 & 5/3 and one to the right of 5/3 . Which of these fulfill the inequality?
     
  5. Sep 17, 2012 #4
    Oh I see, [itex]\frac{2}{3}<x<\frac{5}{3}[/itex]

    Can you help me with a and b?

    Thank you
     
  6. Sep 17, 2012 #5

    SammyS

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    Can you follow the method I used for part (c) ?

    Post your efforts for (a) & (b).

    (There are other methods, of course!)

    Is there a typo in your posted expression for (a)?

    [itex]\displaystyle \frac{1}{2}<\frac{5}{3x+42}<\frac{3}{2}[/itex]

    or

    [itex]\displaystyle \frac{1}{2}<\frac{5}{3x+4}<\frac{3}{2}[/itex]
     
  7. Sep 18, 2012 #6
    Sorry, c) is [itex]\frac{3}{2x-1}<1[/itex]

    So I know that x must be less than 1/2 for which the inequality. In the denominator of 3/2x-1 "I superimpose" if x takes values ​​(-infinity, 1/2) x is negative, thus passing the denominator to multiply with 1 if you change the "<" by "> ".
    Then 3> 2x-1, 4> 2x and 2> x

    Why is x>2? and not 2>x, What am I doing wrong?
     
  8. Sep 18, 2012 #7

    SammyS

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    Multiplying an inequality by a negative number reverses the direction of the inequality.


    If x < 1/2, then 3/(2x-1) is negative, so it's definitely less than 1.

    What if [itex]\displaystyle 2x-1>0\ \text{ i.e. }\ x>\frac{1}{2}\ ?[/itex]

    Then multiplying [itex]\displaystyle \frac{3}{2x-1}<1[/itex] by (2x-1) gives:

    [itex]\displaystyle 3< 2x-1[/itex]

    Solving for x gives: [itex]\displaystyle x>\frac{3}{2}\ .[/itex]

    Put the two answers together.
     
  9. Sep 19, 2012 #8
    I already understand c), but I still don't understand a) and b)
     
  10. Sep 19, 2012 #9

    SammyS

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    Is there a typo in your posted expression for (a)?

    [itex]\displaystyle \frac{1}{2}<\frac{5}{3x+42}<\frac{3}{2}[/itex]

    or

    [itex]\displaystyle \frac{1}{2}<\frac{5}{3x+4}<\frac{3}{2}[/itex]

    or

    something else?
     
  11. Sep 20, 2012 #10
    not only this, but I already know to solve this, thanks anyway
     
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