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Math Help - Writing Algebraic Expression as a Single Fraction

  1. #1
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    Writing Algebraic Expression as a Single Fraction

    Need Help writing the following expression as a single fraction:

    \displaystyle \frac {10}{2x^2-3x-2}-\frac {2}{x-2}

    I've tried a few different ways but not getting near the answer in the book :?

    Any help would be great!

    Cheers

    BIOS
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  2. #2
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    Factor: 2x^2-3x-2=(2x+1)(x-2).
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  3. #3
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    Thanks for the reply Plato. I've tried that bit i think i'm messing up the step after that.

    So say i have:

    \frac {10}{(2x-1)(x-2)}-\frac {2}{x-2}

    After that do i multiply like this:

    \frac {10}{(2x-1)(x-2)}-\frac {2(2x+1)}{(x-2)(2x+1)}
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  4. #4
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    Thanks for the reply Plato. I've tried that bit i think i'm messing up the step after that.

    So say i have:

    \frac {10}{(2x-1)(x-2)}-\frac {2}{x-2}

    After that do i multiply like this:

    \displaystyle \frac {10}{(2x-1)(x-2)}-\frac {2(2x-1)}{(x-2)(2x-1)}
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  5. #5
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    Quote Originally Posted by BIOS View Post
    Thanks for the reply Plato. I've tried that bit i think i'm messing up the step after that.

    So say i have:

    \frac {10}{(2x-1)(x-2)}-\frac {2}{x-2}

    After that do i multiply like this:

    \frac {10}{(2x-1)(x-2)}-\frac {2(2x+1)}{(x-2)(2x+1)}
    It should be \dfrac {10}{(2x+1)(x-2)}-\dfrac {2(2x+1)}{(2x+1)(x-2)}
    Last edited by Plato; September 28th 2010 at 08:22 AM.
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  6. #6
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    Yeah sorry edited that post right after i posted it! Okay so from here where do i go? Haven't done one like this before so unsure what exactly is the best method.
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  7. #7
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    Please reread my reply. I edited it.
    It is correct now.
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  8. #8
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    Okay yeah i can see i mistyped the factoring in the first denominator in the original post (the one you quoted). Still don't know where to go from there however. If i could see it done once that would be very helpful
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  9. #9
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    Do i do this and then cancel:

    \displaystyle \frac {10-2(2x+1)}{(2x+1)(x-2)}

    ?
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  10. #10
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    Quote Originally Posted by BIOS View Post
    Do i do this and then cancel:
    \displaystyle \frac {10-2(2x+1)}{(2x+1)(x-2)}
    Write it as \displaystyle \frac {8-4x}{(2x+1)(x-2)}
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  11. #11
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    So once i have:

    \displaystyle  \frac {8-4x}{(2x+1)(x-2)}

    Do i look for a common factor ala:

    \displaystyle \frac {8-4x}{1} * \frac {1}{(2x+1)(x-2)}<br />
    The answer in the book is:

    \displaystyle \frac {-4}{2x+1}

    Which i can't see how to get.

    If i divide x -2 into the numerator i don't get the numerator in the books answer?

    Obviously i'm missing something! Thanks for your help and patience so far.
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  12. #12
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    Factorise the top:

    \displaystyle \frac {8-4x}{(2x+1)(x-2)}

    =\displaystyle \frac {4(2-x)}{(2x+1)(x-2)}

    Now, take out a factor of -1 from the numerator:

    =\displaystyle \frac {-4(-2+x)}{(2x+1)(x-2)}

    =\displaystyle \frac {-4(x-2)}{(2x+1)(x-2)}

    And I'm sure you can finish.
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  13. #13
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    Thanks Quacky! I really needed that explained. Should be able to get through to rest of these now
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