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Thread: Common denominator

  1. #1
    Member Mollier's Avatar
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    Common denominator

    Hi, my algebra is quite rusty and I can't figure this one out.

    $\displaystyle c_3 = \frac{y_3-y_1}{(x_3-x_1)(x_3-x_2)} - \frac{y_2-y_2}{(x_3-x_2)(x_2-x_1)}
    $
    My book says that "by taking a common denominator we get",

    $\displaystyle c_3 = \frac{\frac{y_3-y_2}{x_3-x_2}-\frac{y_2-y_2}{x_2-x_2}}{x_3-x_1} $

    If I was to find a common denominator for this, it would look something like this,

    $\displaystyle c_3 = c_3 = \frac{y_3-y_1(x_2-x_1)}{(x_3-x_1)(x_3-x_2)(x_2-x_1} - \frac{y_2-y_2(x_3-x_1)}{(x_3-x_2)(x_2-x_1)(x_3-x_1)}$

    I've played with it for a while, but can't seem to arrive at the right expression..

    Any tips besides finding another hobby?
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  2. #2
    Senior Member Educated's Avatar
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    Your book is wrong.

    $\displaystyle c_3 = \dfrac{\frac{y_3-y_2}{x_3-x_2}-\frac{y_2-y_2}{x_2-x_2}}{x_3-x_1}$

    Now in the part where is says $\displaystyle x_2 - x_2$ then it will equal 0. And you know that anything divided by 0 is undefined.


    My answer:


    $\displaystyle c_3 = \dfrac{y_3-y_1}{(x_3-x_1)(x_3-x_2)} - \dfrac{y_2-y_2}{(x_3-x_2)(x_2-x_1)}$

    Now we multiply everything by $\displaystyle (x_3-x_2)$ giving us:

    $\displaystyle c_3(x_3-x_2) = \dfrac{(y_3-y_1)(x_3-x_2)}{(x_3-x_1)(x_3-x_2)} - \dfrac{(y_2-y_2)(x_3-x_2)}{(x_3-x_2)(x_2-x_1)}$

    Cancel out like terms leaving:

    $\displaystyle c_3(x_3-x_2) = \dfrac{y_3-y_1}{x_3-x_1} - \dfrac{y_2-y_2}{x_2-x_1}$

    And then divide everything by $\displaystyle (x_3-x_2)$ to get the answer:

    $\displaystyle c_3 = \dfrac{\frac{y_3-y_1}{x_3-x_1} - \frac{y_2-y_2}{x_2-x_1}}{x_3-x_2}$

    Now we see the part where it says $\displaystyle \frac{y_2-y_2}{x_2-x_1}$ we know that it will be 0 because $\displaystyle y_2-y_2$ = 0. (I should have cancelled that out from the start...)

    This leaves:

    $\displaystyle c_3 = \dfrac{\frac{y_3-y_1}{x_3-x_1}}{x_3-x_2}$

    $\displaystyle c_3 = \dfrac{y_3-y_1}{(x_3-x_1)(x_3-x_2)}$

    Can you carry on from there? Or am I wrong?
    Last edited by Educated; Aug 30th 2010 at 09:47 PM.
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  3. #3
    Member Mollier's Avatar
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    I'm really sorry, I made a typo. It is supposed to be: $\displaystyle y_2-y_2$ would not make much sense I guess.
    Thanks though!
    $\displaystyle c_3 = \frac{y_3-y_1}{(x_3-x_1)(x_3-x_2)} - \frac{y_2-y_1}{(x_3-x_2)(x_2-x_1)}
    $
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  4. #4
    Member Mollier's Avatar
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    The first nominator should be $\displaystyle y_3-y_2$. This book has a lot of errors. Should have seen it at once, but so many terms are giving me a headache.
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  5. #5
    Senior Member Educated's Avatar
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    So from what I gather, this is the formula:

    $\displaystyle c_3 = \dfrac{y_3-y_1}{(x_3-x_1)(x_3-x_2)} - \dfrac{y_3-y_2}{(x_3-x_2)(x_2-x_1)}$


    If that's the case then it would simplify to:

    $\displaystyle c_3 = \dfrac{\frac{y_3-y_1}{x_3-x_1} - \frac{y_3-y_2}{x_2-x_1}}{x_3-x_2}$
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