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Math Help - Common denominator trick?

  1. #1
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    Common denominator trick?

    It's on the tip of my tongue and I can't remember a trick I remember learning to solve these.

     <br />
\frac{3}{(x+2b)(x-b)} + \frac{2}{(3b-x)(x+2b)} + \frac{1}{(x-3b)(b-x)}<br />

    For some reason i feel like there was a trick where you switch the signs and cross multiplied or something like that to get the denominators the same...but i cant remember

    or do i just have to do it the long way and multiply a million times

    Thanks
    Last edited by silencecloak; May 7th 2008 at 10:44 PM.
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  2. #2
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    Just looking at the last term:

    \frac{1}{(x-3b)(b-x)}

    =   \frac{1}{(-1)(3b-x)(x-b)}

    The trick was to factorise -1 out of the brackets.

    If the first term's denominator has (x+2b) instead of (x+2) (is it a typo?), it would have been such a nice fraction...
    Last edited by Gusbob; May 7th 2008 at 10:52 PM.
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  3. #3
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    Quote Originally Posted by Gusbob View Post
    Just looking at the last term:

    \frac{1}{(x-3b)(b-x)}

    =  - \frac{1}{(-1)(3b-x)(x-b)}

    The trick was to factorise -1 out of the brackets.

    If the first term's denominator has (x+2b) instead of (x+2) (is it a typo?), it would have been such a nice fraction...
    yes its a typo


    thanks :P

    so where does that (-1) go now?

    <br />
\frac{7b-x}{(x+2b)(x-b)(3b-x)} + \frac {1}{(-1)(3b-x)(x-b)}<br />

    i can see how this would turn into

    <br />
\frac{(7b-x) + (x+2b)}{(x+2b)(x-b)(3b-x)} <br />

    but i dunno where i put that -1
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  4. #4
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    \frac{3}{(x+2b)(x-b)} + \frac{2}{(3b-x)(x+2b)} + \frac{1}{(x-3b)(b-x)}

     = \dfrac{3(x-3b)}{(x+2b)(x-b)(x-3b)} - \dfrac{2(x-b)}{(x-3b)(x+2b)(x-b)} - \dfrac{x+2b}{(x-3b)(x-b)(x+2b)}

     = \frac{3(x-3b)-2(x-b)-(x+2b)}{(x+2b)(x-b)(x-3b)}

     = \frac{(3x -2x -x)+ (-9b +2b-2b)}{(x+2b)(x-b)(x-3b)}

     = -\frac{9b}{(x+2b)(x-b)(x-3b)}
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  5. #5
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    Quote Originally Posted by silencecloak View Post
    yes its a typo


    thanks :P

    so where does that (-1) go now?

    <br />
\frac{7b-x}{(x+2b)(x-b)(3b-x)} + \frac {1}{(-1)(3b-x)(x-b)}<br />

    i can see how this would turn into

    <br />
\frac{(7b-x) + (x+2b)}{(x+2b)(x-b)(3b-x)} <br />

    but i dunno where i put that -1
    Just so you know, I accidentally put a minus sign in front of my original reply, its not meant to be there.

    I would suggest you do this:

    \frac{7b-x}{(x+2b)(x-b)(3b-x)} - \frac {1}{(3b-x)(x-b)}
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  6. #6
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    Quote Originally Posted by Isomorphism View Post
    \frac{3}{(x+2b)(x-b)} + \frac{2}{(3b-x)(x+2b)} + \frac{1}{(x-3b)(b-x)}

     = \dfrac{3(x-3b)}{(x+2b)(x-b)(x-3b)} - \dfrac{2(x-b)}{(x-3b)(x+2b)(x-b)} - \dfrac{x+2b}{(x-3b)(x-b)(x+2b)}

     = \frac{3(x-3b)-2(x-b)-(x+2b)}{(x+2b)(x-b)(x-3b)}

     = \frac{(3x -2x -x)+ (-9b +2b-2b)}{(x+2b)(x-b)(x-3b)}

     = -\frac{9b}{(x+2b)(x-b)(x-3b)}
    how can you just take the  3b-x and  b-x
    and just flip the signs?
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  7. #7
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    Quote Originally Posted by silencecloak View Post
    how can you just take the  3b-x and  b-x
    and just flip the signs?
    Isnt -(3b-x) = x-3b?

    I will show what I did for the middle one...

    \frac{2}{(3b-x)(x+2b)} = \frac{2}{(3b-x)(x+2b)} {\color{red}\times \frac{-(x-b)}{-(x-b)}}

    \frac{2}{(3b-x)(x+2b)} {\color{red}\times \frac{-(x-b)}{-(x-b)}} = \frac{-2(x-b)}{-(3b-x)(x+2b)(x-b)}

    Now use -(3b-x) = x-3b

    \frac{-2(x-b)}{-(3b-x)(x+2b)(x-b)} = \frac{-2(x-b)}{(x-3b)(x+2b)(x-b)}

    Understood?
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    Quote Originally Posted by Isomorphism View Post
    Isnt -(3b-x) = x-3b?

    I will show what I did for the middle one...

    \frac{2}{(3b-x)(x+2b)} = \frac{2}{(3b-x)(x+2b)} {\color{red}\times \frac{-(x-b)}{-(x-b)}}

    \frac{2}{(3b-x)(x+2b)} {\color{red}\times \frac{-(x-b)}{-(x-b)}} = \frac{-2(x-b)}{-(3b-x)(x+2b)(x-b)}

    Now use -(3b-x) = x-3b

    \frac{-2(x-b)}{-(3b-x)(x+2b)(x-b)} = \frac{-2(x-b)}{(x-3b)(x+2b)(x-b)}

    Understood?
    Yes, but isn't
    \frac{(3x -2x -x)+ (-9b +2b-2b)}{(x+2b)(x-b)(x-3b)}<br />

    -9b not 9b ?

    or does that negative come back from somewhere
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  9. #9
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    Quote Originally Posted by silencecloak View Post
    Yes, but isn't
    \frac{(3x -2x -x)+ (-9b +2b-2b)}{(x+2b)(x-b)(x-3b)}<br />

    -9b not 9b ?

    or does that negative come back from somewhere
    Its there, I marked it in red here:
    = {\color{red} -}\frac{9b}{(x+2b)(x-b)(x-3b)}
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  10. #10
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    Quote Originally Posted by Isomorphism View Post
    Its there, I marked it in red here:
    = {\color{red} -}\frac{9b}{(x+2b)(x-b)(x-3b)}
    Ah, my apologies.

    Thanks for stickin with me.

    Been a couple years since I have done all this and I have to pass a test to get into calculus!

    Hmm one more thing

    Would it be wrong to just go  3(3b-x) + 2(x-b) + (x+2b)

    all over the LCD, this comes out to 9b perfectly, but I have a feeling this is an isolated case where this works.
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  11. #11
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    Quote Originally Posted by silencecloak View Post
    Would it be wrong to just go  3(3b-x) + 2(x-b) + (x+2b)

    all over the LCD, this comes out to 9b perfectly, but I have a feeling this is an isolated case where this works.
    Whether stuff cancels out is purely problem dependent... And what you said is not wrong
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