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Math Help - Adding/Subracting Fractions with Variables

  1. #1
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    Adding/Subracting Fractions with Variables

    I need some help understanding whats going on here.

    Description:
    When adding fractions with variables in one or more denominators, the LCD will have each variable to its highest power as a factor.
    \frac{13}{xy^2}-\frac{6}{yz} = \frac{13}{xy^2}*\frac{z}{z}-\frac{6}{yz}*\frac{xy}{xy}=\frac{13z}{xy^2z}-\frac{6xy}{xy^2z} =\frac{13z-6xy}{xy^2z}

    Can someone please help me through this process, I understand picking out the highest exponent of each variable but after that its unclear. Also how do you know to arrange the expression? For instance \frac{13}{xy^2}*\frac{z}{z} How would I know that the \frac{z}{z} gets matched up with the \frac{13}{xy^2} and not on the other end of the expression with the \frac{6xy}{xy^2z}?

    Many Thanks
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    Quote Originally Posted by cmf0106 View Post
    I need some help understanding whats going on here.

    Description:
    When adding fractions with variables in one or more denominators, the LCD will have each variable to its highest power as a factor.
    \frac{13}{xy^2}-\frac{6}{yz} = \frac{13}{xy^2}*\frac{z}{z}-\frac{6}{yz}*\frac{xy}{xy}=\frac{13z}{xy^2z}-\frac{6xy}{xy^2z} =\frac{13z-6xy}{xy^2z}

    Can someone please help me through this process, I understand picking out the highest exponent of each variable but after that its unclear. Also how do you know to arrange the expression? For instance \frac{13}{xy^2}*\frac{z}{z} How would I know that the \frac{z}{z} gets matched up with the \frac{13}{xy^2} and not on the other end of the expression with the \frac{6xy}{xy^2z}?

    Many Thanks
    Let me introduce a whole different approach. If you don't like it, you can still pursue the one that you started here.

    Your LCD is xy^2z (all the different factors of the denominator with the highest exponent)

    Multiply that LCD times each numerator and simplify. Apply the results over the LCD of xy^2z

    \frac{13}{xy^2}-\frac{6}{yz}

    \frac{xy^2z(\frac{13}{xy^2})-xy^2z(\frac{6}{yz})}{xy^2z}

    Now simplify the numerator:

    \frac{13z-6xy}{xy^2z}
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  3. #3
    Member cmf0106's Avatar
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    Thanks Masters! Your advice makes it very clear. I assume this will always work, and I can continue to build on this method even with more advance problems. I cannot thank you enough the concept in the book makes next to no sense to me. Thank you!!!!
    Last edited by cmf0106; August 15th 2008 at 09:46 AM.
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    Masters, would you please show your work on how to simplify this problem?Thanks.
    \frac{xy^2z(\frac{13}{xy^2})-xy^2z(\frac{6}{yz})}{xy^2z}<br />
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    \frac{{\color{red}x}{\color{blue}y^2}z(\frac{13}{{  \color{red}x}{\color{blue}y^2}}) \: - \: x{\color{blue}y^{2}}{\color{magenta}z}(\frac{6}{{\  color{blue}y}{\color{magenta}z}})}{xy^2z}

    Notice that the variables colour-coded can be cancelled out (basically, dividing a number by itself gives you 1 which you don't need to write out).

    When dividing powers in the case of {\color{blue}y^2}, remember that: \frac{y^{n}}{y^{m}} = y^{n-m}. Here, n = 2 and m = 1.
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  6. #6
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    Quote Originally Posted by allyourbass2212 View Post
    Masters, would you please show your work on how to simplify this problem?Thanks.
    \frac{xy^2z(\frac{13}{xy^2})-xy^2z(\frac{6}{yz})}{xy^2z}
    \frac{xy^2z(\frac{13}{xy^2})-xy^2z(\frac{6}{yz})}{xy^2z}= \frac{{\frac{13xy^2z}{xy^2}} - \frac{6xy^2z}{yz}}{xy^2z}

    In the numerator of the complex fraction above, you can simplify by dividing out the variables resulting in:

    \frac{13z-6xy}{xy^2z}
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    This is going to be a real stupid question but in the final answer
    \frac{13}{xy^2}-\frac{6}{yz} =<br />
\frac{13}{xy^2}*\frac{z}{z}-\frac{6}{yz}*\frac{xy}{xy}=\frac{13z}{xy^2z}-\frac{6xy}{xy^2z}<br />
=\frac{13z-6xy}{xy^2z}<br />

    Why does it not simply even further? IE There is a single x & y variable in the numerator and denominator, how come those cannot be canceled out and have the answer read \frac{13-6y}{y^2}?
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    Quote Originally Posted by allyourbass2212 View Post
    This is going to be a real stupid question but in the final answer
    \frac{13}{xy^2}-\frac{6}{yz} =<br />
\frac{13}{xy^2}*\frac{z}{z}-\frac{6}{yz}*\frac{xy}{xy}=\frac{13z}{xy^2z}-\frac{6xy}{xy^2z}<br />
=\frac{13z-6xy}{xy^2z}<br />

    Why does it not simply even further? IE There is a single x & y variable in the numerator and denominator, how come those cannot be canceled out and have the answer read \frac{13-6y}{y^2}?
    You can only "cancel" across factors, not terms that have been added or subtracted. For instance,

    \frac{xy^2z}{x^2y}=\frac{yz}{x} with numerator and denominator being all factors (no + or - signs)
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    So you can only "cancel" out that way whenever the numerator and denominator numbers are connected by multiplication, if there is a + or - in the numerator you cannot cancel out that way.

    \frac{13z-6xy}{xy^2z}<br />

    So in this instance since there is a '-' in the numerator this is as simplified as its going to get.

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    Quote Originally Posted by cmf0106 View Post
    I need some help understanding whats going on here.

    Description:
    When adding fractions with variables in one or more denominators, the LCD will have each variable to its highest power as a factor.
    \frac{13}{xy^2}-\frac{6}{yz} = \frac{13}{xy^2}*\frac{z}{z}-\frac{6}{yz}*\frac{xy}{xy}=\frac{13z}{xy^2z}-\frac{6xy}{xy^2z} =\frac{13z-6xy}{xy^2z}

    Can someone please help me through this process, I understand picking out the highest exponent of each variable but after that its unclear. Also how do you know to arrange the expression? For instance \frac{13}{xy^2}*\frac{z}{z} How would I know that the \frac{z}{z} gets matched up with the \frac{13}{xy^2} and not on the other end of the expression with the \frac{6xy}{xy^2z}?

    Many Thanks
    you should follow the PEMDAS rule
    Parenthesis, Exponent, Multiplication, Division, Addition then Subtraction
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