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Math Help - hard fraction!

  1. #16
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    Quote Originally Posted by harish21 View Post
    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.




    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!
    okay i think i get it now, so would 0=6x^2 -6x be the answer?
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  2. #17
    MHF Contributor harish21's Avatar
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    Does your question tell you to find the value of x?
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  3. #18
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    usually with questions like these i usually just find X via quadratic equation or by dividing a number by the amount of X's.

    how would i work out x with this question?
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  4. #19
    MHF Contributor harish21's Avatar
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    You have:

    0=6x^2-6x

    Find x from here..
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  5. #20
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    would i square root 6x^2=6
    0=6-6

    plus 1 to each side to get:
    6=6
    6/6=1
    x=1?
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  6. #21
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    Quote Originally Posted by andyboy179 View Post
    would i square root 6x^2=6
    0=6-6

    plus 1 to each side to get:
    6=6
    6/6=1
    x=1?
    No, to find all answers, write as a multiplication of factors...

    6x^2-6x=0

    6xx-6x=0

    (x)6x+(-1)6x=0

    6x is common to both, so factor...

    6x(x-1)=0

    6 is not 0, so x(x-1)=0

    Those factors now tell you the answers because any multiple of zero is zero.
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  7. #22
    MHF Contributor Unknown008's Avatar
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    Yes, x = 1 but you forgot 0

    If you so much want to use the formula, you can make this:

    0 = ax^2 + bx + c

    x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

    Here, you have 0 = 6x^2 - 6x

    So, a = 6, b = -6, c = 0
    Last edited by Unknown008; February 19th 2011 at 01:28 AM. Reason: typo
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  8. #23
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    So by doing the quadratic equation I would end with the answers of 1 and 0 for x??
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  9. #24
    MHF Contributor Unknown008's Avatar
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    Did you get something else than 1 and 0?
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  10. #25
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    No I did the quadratic and get 1 and 0.
    To get it I did:
    X=6+ (square root)36-4x6x0/12=1
    X=6- (square root)36-4x6x0/12=0
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