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Math Help - fractions with negative exponents

  1. #16
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    Just a quick one what would happen it that was in the denominator?

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  2. #17
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    Quote Originally Posted by anthonye View Post
    Just a quick one what would happen it that was in the denominator?

    Good night.
    then you cannot split it apart. sleep well.
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  3. #18
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    Then how would you solve it please post example.

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  4. #19
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    Can someone please show an example.
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  5. #20
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    In that situation it generally can't be simplified very much. Here's my random example.

    \frac{x^{-1}y^{-3}z^{2}}{x^4y^{-2}+7}

    =\displaystyle\frac{\frac{1}{x}\times\frac{1}{y^3}  \times z^2}{x^4\frac{1}{y^2}+7}

    Now consider the numerator:
    =\displaystyle{\frac{1}{x}\times\frac{1}{y^3}\time  s z^2}

    =\displaystyle\frac{z^2}{xy^3}

    The denominator:

    \displaystyle x^4\frac{1}{y^2}+7

    =\displaystyle\frac{x^4}{y^2}+7

    =\displaystyle\frac{x^4+7y^2}{y^2}

    So I could rewrite my example as:

    \displaystyle\frac{\frac{z^3}{xy^3}}{\frac{x^4+7y^  2}{y^2}}
    <br />
=\displaystyle\frac{z^3}{xy^3}\times\frac{y^2}{x^4  +7y^2}

    And you'd carry on simplifying from there. I don't know how standard this method is, though I would assume it can be applied to most, if not all, situations.
    Last edited by Quacky; March 1st 2011 at 11:47 AM.
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  6. #21
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    Unfortunately
    going from 6 terms to 7 terms
    wouldn't often be classed as a "simplification".
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  7. #22
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    Quote Originally Posted by Quacky View Post
    In that situation it generally can't be simplified very much. Here's my random example.

    \frac{x^{-1}y^{-3}z^{2}}{x^4y^{-2}+7}

    =\displaystyle\frac{\frac{1}{x}\times\frac{1}{y^3}  \times z^2}{x^4\frac{1}{y^2}+7}

    Now consider the numerator:
    =\displaystyle{\frac{1}{x}\times\frac{1}{y^3}\time  s z^2}

    =\displaystyle\frac{z^2}{xy^3}

    The denominator:

    \displaystyle x^4\frac{1}{y^2}+7

    =\displaystyle\frac{x^4}{y^2}+7

    =\displaystyle\frac{x^4+7y^2}{y^2}

    So I could rewrite my example as:

    \displaystyle\frac{\frac{z^3}{xy^3}}{\frac{x^4+7y^  2}{y^2}}
    <br />
=\displaystyle\frac{z^3}{xy^3}\times\frac{y^2}{x^4  +7y^2}

    And you'd carry on simplifying from there. I don't know how standard this method is, though I would assume it can be applied to most, if not all, situations.
    Quote Originally Posted by Archie Meade View Post
    Unfortunately
    going from 6 terms to 7 terms
    wouldn't often be classed as a "simplification".
    Haha, that's because I didn't finish. I hope you're teasing me. If not, then how are you defining 'term'? I don't see how you've identified 7.

    =\displaystyle\frac{z^3y^2}{x^5y^3+7xy^5}

    =\displaystyle\frac{z^3}{x^5y+7xy^3}

    If I've understood you correctly, then by your definition there are still 6 "terms" here, and whilst it might not be mathematically any simpler, I know which I'd rather use! I would argue that there are just three terms in my final answer and three terms in my initial expression, but regardless, whilst the amount of terms stays the same, there are no negative indices in my final answer. Also, by your definition of a term, Skeeter's previous example also fails to be "simplified," unless I'm misinterpreting what you mean. There was a thread in the chatroom about this, actually - 'term' is a very ambiguous word. But I'm waffling: my point is that whilst it might be mathematically indifferent, I acknowledged that there was little that could be done in the way of simplification, and perhaps my method is not classed as 'simplification' as you would define it, but you can ask anyone which they'd rather work with, and I think you can probably predict the most common answer.

    Edit: Furthermore, the OP was asking how you'd solve such a fraction. I think that by writing it in this form, solving for any of the variables is a lot simpler than it was in the original mess.
    Last edited by Quacky; March 1st 2011 at 03:13 PM.
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  8. #23
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    Quote Originally Posted by Quacky View Post
    In that situation it generally can't be simplified very much. Here's my random example.

    \dfrac{x^{-1}y^{-3}z^{2}}{x^4y^{-2}+7}
    \dfrac{x^{-1}y^{-3}z^{2}}{x^4y^{-2}+7} \cdot \dfrac{xy^3}{xy^3} = \dfrac{z^2}{x^5y+7xy^3}
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  9. #24
    Super Member Quacky's Avatar
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    Red face

    Quote Originally Posted by skeeter View Post
    \dfrac{x^{-1}y^{-3}z^{2}}{x^4y^{-2}+7} \cdot \dfrac{xy^3}{xy^3} = \dfrac{z^2}{x^5y+7xy^3}
    . I knew that.

    Anyway, my point was that it couldn't be simplified very much, but that got rid of the negative exponents. Is that not what the OP was asking?
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