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Homework Help: Q2 - Inequality with fractions

  1. Aug 23, 2010 #1
    1. The problem statement, all variables and given/known data

    2 - ((x-3)/(x-2)) ≥ ((x-5)/(x-1))

    2. Relevant equations



    3. The attempt at a solution

    I just want to make sure that certain operations that are allowed with fractions in equations are still valid (or not) in inequalities.

    2 - ((x-3)/(x-2)) ≥ ((x-5)/(x-1))

    Can I remove the fractions by multiplying by the LCM here?

    When I do, I get to the solution:

    2(x-2)(x-1) - (x-3)(x-1) ≥ (x-5)(x-2)

    this implies

    x2-2x +1 ≥ x2-7x+10
    -2x+1 ≥ -7x+10
    x ≥ 9/5

    However, when I put 9/5 into the original inequality I get:

    -2 ≥ -4

    Of course this is right, however, I was expecting the same values on each side. Is there something wrong with this expectation, or something wrong with my calculation??

    Thanks in advance,
     
  2. jcsd
  3. Aug 23, 2010 #2
    Check your plug in to the original equation. I get -4 for both sides.
     
  4. Aug 23, 2010 #3
    Yes you are right! That is good news...been at it too long for me still being able to read my own handwriting or able to do simple head calculations :S

    Thanks,
     
  5. Aug 23, 2010 #4

    hunt_mat

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    Sorry but you can't do that, what if x is between 1 and 2?, this will change the sign of the inequality, you can only multiply by positive numbers if you want to keep the inequality the same., you can however multiply by (x-1)^{2}(x-1)^{2} and not change the sign.
     
  6. Aug 23, 2010 #5
    Hmm, I thought something like that might be the case.

    So you are saying that the Lowest Common Multiple that is in this case allowed is (x-2)2(x-1)2

    Is this then the proper way to solve this?

    2(x-2)2(x-1)2 - (x-3)(x-2)(x-1)2 ≥ (x-5)(x-1)(x-2)2 looks a bit like a mess:

    2(x2-4x+4)(x2-2x+1) - (x2-5x+6)(x2-2x+1) ≥ (x2-6x+5)(x2-4x+4)
     
  7. Aug 23, 2010 #6

    hunt_mat

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    It does look a mess but there is a factor of (x-2)(x-1), so the total factorisation is:
    [tex]
    (x-1)(x-2)(5x-9)\geqslant 0
    [/tex]
    Now you can examine the inequality at your leisure.
     
  8. Aug 23, 2010 #7
    I'm starting to feel rather confused even though I thought I had grasped all the basic ideas (like taking the common factor).

    I have no idea how you just came up with that last inequality from the previous one with all the exponents in it. Even after rereading my sources for this type of problem I am at a loss...

    Do I first work out all the equations and than factor back? Or do I start with the factoring right away, while the (x-2)(x-1) factor is still visible?

    Something I could imagine would be:

    2*2-(x-3)(x-1) ≥ (x-5)(x-2) after dividing all terms by the same factor (x-2)(x-1) but I have a feeling this is not the right way to go...
     
  9. Aug 23, 2010 #8

    hunt_mat

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    Okay, you correctly wrote down:
    [tex]
    2(x-2)^{2}(x-1)^{2} - (x-3)(x-2)(x-1)^{2} \geqslant (x-5)(x-1)(x-2)^{2}
    [/tex]
    All I did was take everything over to the LHS
    [tex]
    2(x-2)^{2}(x-1)^{2} - (x-3)(x-2)(x-1)^{2}-(x-5)(x-1)(x-2)^{2} \geqslant 0
    [/tex]
    and take out a factor of (x-1)(x-2)
    [tex]
    (x-1)(x-2)[2(x-2)(x-1)- (x-3)(x-1)-(x-5)(x-2)] \geqslant 0
    [/tex]
    and expand the square bracket to obtain what I did.

    Mat
     
  10. Aug 23, 2010 #9
    Ha, I should have checked your actual work instead of grabbing at the low hanging fruit. :)
     
  11. Aug 24, 2010 #10
    Just to make sure I understand inequalities fully:

    x: [1,(9/5)]U[2, infinity)

    Is this correct?
     
  12. Aug 24, 2010 #11

    hunt_mat

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    [tex]
    x\geqslant 2
    [/tex]
     
  13. Aug 25, 2010 #12

    ehild

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    x can not be either 1 or 2, so I would say x: (1,(9/5)]U(2, infinity), but it is correct otherwise.

    ehild
     
  14. Aug 25, 2010 #13
    Ok, so it seems I do not quite understand it yet.

    My reasoning was the following:

    If x = 2, x =1 and x = 9/5 the equation is equal to 0 (0 times something is 0). Why is this not correct in this case?

    Also, when two of the three factors are minus and one is positive, this generates a positive number right? Thus x may be between 1 and 9/5 as well, meaning both the (x-2) and (5x-9) terms are negative but the (x-1) positive.

    Ibviously when x is greater than 2, all three factors are positive and this generates a positive number.

    Also, I got two different answers :) Can you elaborate please?
     
  15. Aug 25, 2010 #14

    ehild

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    You need to fulfil the original inequality, which loses sense if x=2 or x=1 because of the zero denominators.
    And yes, you have two domains for x to fulfil the original inequality.

    If you plot the functions at both sides you will see the the domains where the left side is greater or equal to the right side, and also the points where the functions change between -infinity and infinity.

    ehild
     
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