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Math Help - hard fraction!

  1. #1
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    hard fraction!

    hi,
    i need to work out this fraction:

    3/2x-1 - 4/3x-1

    i can do the first part and worked it out and you get a quadratic equation as shown in my picture:
    hard fraction!-quadratic.jpg

    when i type in to my calculator 36-4x6x6 it =-108?

    -108 doesn't work!!! why is it coming up with this? have i started my quadratic equation wrong?

    help would be much appreciated!!!!

    thanks!
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  2. #2
    MHF Contributor Unknown008's Avatar
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    If you need to work out, then you only need to simplify. There are no equal signs and hence, you cannot solve for x.

    \dfrac{3}{2x - 1} - \dfrac{4}{3x-1} = \dfrac{3(3x-1) - 4(2x-1)}{(2x-1)(3x-1)}

    = \dfrac{9x - 3 - 8x + 4}{6x^2 - 2x - 3x + 1}

    = \dfrac{x+1}{6x^2 - 5x + 1}

    That's about all you can do...
    Last edited by Unknown008; February 18th 2011 at 10:53 AM. Reason: Distributive mistake on numerator corrected.
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  3. #3
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    Quote Originally Posted by andyboy179 View Post
    hi,
    i need to work out this fraction:

    3/2x-1 - 4/3x-1

    i can do the first part and worked it out and you get a quadratic equation as shown in my picture:
    Click image for larger version. 

Name:	quadratic.jpg 
Views:	20 
Size:	34.6 KB 
ID:	20861

    when i type in to my calculator 36-4x6x6 it =-108?

    -108 doesn't work!!! why is it coming up with this? have i started my quadratic equation wrong?

    help would be much appreciated!!!!

    thanks!
    Hi andyboy179,

    What do you mean "work out this fraction". What is there to work out? And there is no equation here.

    And, please use grouping symbols (parenthesis) to show us what's in the denominators of your fractions.

    Restate the problem exactly the way it's stated in your book or worksheet or whatever.
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  4. #4
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    Quote Originally Posted by Unknown008 View Post
    If you need to work out, then you only need to simplify. There are no equal signs and hence, you cannot solve for x.

    \dfrac{3}{2x - 1} - \dfrac{4}{3x-1} = \dfrac{3(3x-1) - 4(2x-1)}{(2x-1)(3x-1)}

    = \dfrac{3x - 3 - 8x + 4}{6x^2 - 2x - 3x + 1}

    = \dfrac{1-5x}{6x^2 - 5x + 1}

    That's about all you can do...
    on this part:

    = \dfrac{3x - 3 - 8x + 4}{6x^2 - 2x - 3x + 1}

    how did you get 3x-3 wouldn't it be 9x-3
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  5. #5
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    really really sorry, miss typed my original question!!

    it is:


    \dfrac{3}{2x - 1} - \dfrac{4}{3x-1} =1
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  6. #6
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by andyboy179 View Post
    on this part:

    = \dfrac{3x - 3 - 8x + 4}{6x^2 - 2x - 3x + 1}

    how did you get 3x-3 wouldn't it be 9x-3
    I think Unknown008 made a small mistake. The numerator should be 9x - 3 - 8x + 4

    so, you have:

    \dfrac{9x - 3 - 8x + 4}{6x^2 - 2x - 3x + 1}=1

    can you finish?
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  7. #7
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    Quote Originally Posted by harish21 View Post
    I think Unknown008 made a small mistake. The numerator should be 9x - 3 - 8x + 4

    so, you have:

    \dfrac{9x - 3 - 8x + 4}{6x^2 - 2x - 3x + 1}=1

    can you finish?
    would it go into a quadratic equation and end as 0.6 and 0??
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  8. #8
    MHF Contributor harish21's Avatar
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    It does go into a quadratic equation, but only one of the answers you have written is correct! How did you get 0.83? Was that a guess?If not, Please show your work
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  9. #9
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    sorry i forgot to take an x away from 5x! my answers are 0.6 and 0
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  10. #10
    MHF Contributor Unknown008's Avatar
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    Sorry for that, I'm really tired and got a lot of things to do >.< Thanks harish
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  11. #11
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by andyboy179 View Post
    sorry i forgot to take an x away from 5x! my answers are 0.6 and 0
    NO! Still incorrect. My question is HOW did you get there?! Please show the process!!
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  12. #12
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    hard fraction!-.jpg

    then quadratic equation with answers as 0.6 and 0
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  13. #13
    MHF Contributor harish21's Avatar
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    okay. you subtract x and 1 from both the left hand side and the right hand side, that is,

    6x^2-5x+1=x+1

    6x^2-5x+1-x-1=x+1-x-1

    6x^2-6x+0=0

    6x^2-6x=0

    x=....
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  14. #14
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    Quote Originally Posted by harish21 View Post
    okay. you subtract x and 1 from both the left hand side and the right hand side, that is,

    6x^2-5x+1=x+1

    6x^2-5x+1-x-1=x+1-x-1

    6x^2-6x+0=0

    6x^2-6x=0

    x=....
    i don't understand that because you have:
    X+1=6x^2-5+1
    minus x and 1 from each side leaves you with
    0=6x^2-4+0

    how did you get 6x^2-5x+1=x+1?
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  15. #15
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by andyboy179 View Post
    i don't understand that because you have:
    X+1=6x^2-5+1
    minus x and 1 from each side leaves you with
    0=6x^2-4+0
    No it does not. You are subtracting x and 1 from both sides of the equation.

    (x+1)=(6x^2-5x+1)

    subtract 1 from both the sides:

    (x+1)-1=(6x^2-5x+1)-1

    that leaves you with:

    x = 6x^2-5x

    now you subtract x from both the sides. this gives you:

    x-x=(6x^2-5x)-x

    which gives you:

    0=6x^2-5x-x

    therefore 0=6x^2-6x

    ************************************************** *************
    Note: \boxed{-5x-x=-6x} \;\;and \;\;\boxed{-5x+x=-4x}

    In the first box, you are subtracting x from -5x. In the second box, you are adding x to -5x. This is where your confusion lies.

    If you still dont understand it, please refer to your books/notes/instructor to avoid trouble in the future.


    how did you get 6x^2-5x+1=x+1?
    Please refer to post Number 2 by Unknown008

    your question was \dfrac{3}{2x - 1} - \dfrac{4}{3x-1}=1

    then Unknown008's method (which you've been told multiple number of times) takes you :

     \dfrac{x+1}{6x^2 - 5x + 1}=1

    multiplying both sides of this equation by 6x^2 - 5x + 1.gives you:

    x+1=6x^2 - 5x + 1.

    Have fun!
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