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Math Help - Algebraic fraction

  1. #1
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    Algebraic fraction

    Hi,
    We have to express X in terms of Y.

    The correct answer is x = \frac{5y - 2}{y + 3}

    Multiplying both sides by (5 - X) we get

    5y - xy = 3x + 2; 5y - 2 = 3x + xy; \frac{5y - 2}{3} = x + xy

    I am going to stop there as I am on the wrong track. I would appreciate your help in finding the solution.

    Cheers,
    Sean.
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  2. #2
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    Re: Algebraic fraction

    Quote Originally Posted by Seaniboy View Post
    Hi,
    We have to express X in terms of Y.

    The correct answer is x = \frac{5y - 2}{y + 3}

    Multiplying both sides by (5 - X) we get

    5y - xy = 3x + 2; 5y - 2 = 3x + xy; \frac{5y - 2}{3} = x + xy

    I am going to stop there as I am on the wrong track. I would appreciate your help in finding the solution.

    Cheers,
    Sean.
    What's the original equation?
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  3. #3
    MHF Contributor ebaines's Avatar
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    Re: Algebraic fraction

    From your other posts I assume you started with  y = \frac {3x+2}{5-x}, and want to rearrange to make x the subject. By the way - to avoid confusion iit would have been better to add your attempt to your previous thread rather than start a new one - that way we wouldn't have to guess at what the original problem was.

    Quote Originally Posted by Seaniboy View Post
    Multiplying both sides by (5 - X) we get

    5y - xy = 3x + 2; 5y - 2 = 3x + xy;
    So far so good

    Quote Originally Posted by Seaniboy View Post
    \frac{5y - 2}{3} = x + xy
    No - if you divide both sides by 3 the right hand side becomes  \frac {3x +xy} 3 = x + \frac {xy}3. This is not the right approach. Instead from the previous expression you can gather the x terms:

     5y-2 = 3x + xy = x(3 + y)

    Now divide both sides by  (3+y) and you're done.
    Last edited by ebaines; July 8th 2013 at 09:10 AM.
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  4. #4
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    Re: Algebraic fraction

    Thanks. Yes, I have regularly given myself less than a pass grade in the past.
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  5. #5
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    Re: Algebraic fraction

    Hi,
    Apologies for the confusion and well deduced.
    Thanks once again for your help.
    Cheers,
    Sean.
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  6. #6
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    Re: Algebraic fraction

    can any one help me in this type of sum.. I cn't got it.
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