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Math Help - Need help with two rational expressions

  1. #1
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    Need help with two rational expressions

    I'm having problems with two rational expressions that need to be worked out. I really need a step by step example so that I can do it on the test I'll be having soon.

    is it  \frac {8(y - 4)}{20-y} with y not equal to -3 for the domain?

    \frac {4y - 16}{5y+15} * \frac {2y + 6}{4-y}

    and

    \frac {x - 13}{x^3(3 - x)} * \frac {x(x - 3)}{5}
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  2. #2
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    Quote Originally Posted by Skynt View Post

    \frac {x - 13}{x^3(3 - x)} * \frac {x(x - 3)}{5}
    -\frac{(x-13)}{x^3(x-3)} \cdot \frac{x(x-3)}{5}

     - \frac{(x-13)}{5x^2(x-3)}
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  3. #3
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    Quote Originally Posted by Skynt View Post
    I'm having problems with two rational expressions that need to be worked out. I really need a step by step example so that I can do it on the test I'll be having soon.

    \frac {4y - 16}{5y+15} * \frac {2y + 6}{4-y}
    note that we can factor these:

    \frac {4y - 16}{5y+15} \cdot \frac {2y + 6}{4-y} = \frac {4(y - 4)}{5(y+3)} \cdot \frac {2(y + 3)}{4-y}

    can you continue?


    and

    \frac {x - 13}{x^3(3 - x)} * \frac {x(x - 3)}{5}
    note that \frac {x - 13}{x^3(3 - x)} \cdot \frac {x(x - 3)}{5} = \frac {x - 13}{x^3(3 - x)} \cdot \frac {-x(3 - x)}{5}

    can you continue?
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    I'm quite confused lol, can't you cancel out x-3 (sorry) and end up with
     \frac {-(x-13)}{5x^2} ?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Skynt View Post
    I'm quite confused lol, can't you cancel out x-3 (sorry) and end up with
     \frac {-(x-13)}{5x^2} ?
    yes, that is correct

    Quote Originally Posted by Skynt View Post
    I'm having problems with two rational expressions that need to be worked out. I really need a step by step example so that I can do it on the test I'll be having soon.

    is it  \frac {8(y - 4)}{20-y} with y not equal to -3 for the domain?
    no, it is for y \ne 20, since y = 20 is the problem spot here
    Last edited by ThePerfectHacker; October 16th 2007 at 07:56 PM.
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    Sorry for more questions, but I really need to grasp this completely. Thanks for the help so far, but what if I have:
    \frac {2x^2y}{xy - y} lost on this one.
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    Quote Originally Posted by Skynt View Post
    Sorry for more questions, but I really need to grasp this completely. Thanks for the help so far, but what if I have:
    \frac {2x^2y}{xy - y} lost on this one.
    \frac{2x^2y}{y(x-1)} = \frac{2x^2}{x-1}
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