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Thread: Algebraic Fractions Questions

  1. #1
    Junior Member PythagorasNeophyte's Avatar
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    Algebraic Fractions Questions

    Hello everyone.

    I have two urgent maths questions related to simplifying and solving algebraic fractions. I will be very gratified if you could pinpoint my mistakes and correct them with clear explanations.

    1.$\displaystyle \dfrac{(a-b)^2 +a(a+b) -11b^2}{a^2 -4b^2}$

    2.$\displaystyle (\dfrac{-2}{xy})^2$ X $\displaystyle \dfrac{y(y -x)^2}{4} $ X $\displaystyle \dfrac{x^2y^2}{(x^2 -y^2)}$

    Here are my workings respectively-

    1. $\displaystyle \dfrac{(a-b)^2 +a(a+b) -11b^2}{a^2 -4b^2}$

    =$\displaystyle \dfrac{a^2-2ab+b^2+a^2+ab-11b^2}{(a-2b)(a+2b)}$

    =$\displaystyle \dfrac{a^2+a^2-2ab+ab+b^2-11b^2}{(a-2b)(a+2b)}$

    =$\displaystyle \dfrac{2a^2-ab-10b^2}{(a-2b)(a+2b)}$

    =$\displaystyle \dfrac{(-2a+5b)(-a-2b)}{(a-2b)(a+2b)}$ ????

    2. $\displaystyle (\dfrac{-2}{xy})^2$ X $\displaystyle \dfrac{y(y -x)^2}{4} $ X $\displaystyle \dfrac{x^2y^2}{(x^2 -y^2)}$

    = $\displaystyle \dfrac{-4}{x^2y^2}$ X $\displaystyle \dfrac{y(y^2-2xy+x^2)}{4}$ X $\displaystyle \dfrac{x^2y^2}{(x-y)(x+y}$ ?????
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  2. #2
    Senior Member nikhil's Avatar
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    1) (-2a+5b)(-a-2b)=(2a-5b)(a+2b) now cancel (a+2b) from numerator and denominator then you will get the final answer
    2) hints
    (a/b)x(b/c)=a/c [b get cancel from numerator and denominator]
    (-2/xy)^2=4/(xy)^2
    (y-x)^2=(x-y)^2 use this to cancel (x-y) from denominator and numerator
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  3. #3
    Junior Member PythagorasNeophyte's Avatar
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    Thank you nikhil, however I still could not figure out what you are trying to teach. Can you further elaborate on it?
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  4. #4
    Senior Member Educated's Avatar
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    For the first one:

    $\displaystyle =\dfrac{(-2a+5b)(-a-2b)}{(a-2b)(a+2b)}$

    This is the same as:

    $\displaystyle =\dfrac{-1(-2a+5b)(a+2b)}{(a-2b)(a+2b)}$

    The a + 2b cancel eachother out, leaving:

    $\displaystyle =\dfrac{-1(-2a+5b)}{(a-2b)}$

    $\displaystyle =\dfrac{2a-5b}{a-2b}$



    For the second one:

    $\displaystyle \dfrac{-4}{x^2y^2}\times \dfrac{y(y^2-2xy+x^2)}{4}\times \dfrac{x^2y^2}{(x-y)(x+y)}$


    What nikhil was trying to say was that the x^2 * y^2 cancel eachother out because one is on the numerator and the other is on the denominator, leaving:

    $\displaystyle \dfrac{y(y^2-2xy+x^2)}{4}\times \dfrac{4}{(x-y)(x+y)}~~~~~ = \dfrac{y(x-y)(x-y)}{4}\times \dfrac{4}{(x-y)(x+y)}$


    Note: (-2)^2=4
    Can you carry on from there?
    Last edited by Educated; Aug 30th 2010 at 01:59 AM.
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  5. #5
    Junior Member PythagorasNeophyte's Avatar
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    Wow, that's genuinely quick!

    So, for question 2:

    $\displaystyle
    \dfrac{y(y^2-2xy+x^2)}{4}\times \dfrac{4}{(x-y)(x+y)}~~~~~ = \dfrac{y(x-y)(x-y)}{4}\times \dfrac{4}{(x-y)(x+y)}

    $

    (cancel off the fours and then cancel off $\displaystyle (x-y)$ and $\displaystyle (x-y)$)

    =$\displaystyle \dfrac{y(x-y)}{x+y}$
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  6. #6
    Senior Member Educated's Avatar
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    Seems right to me.
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  7. #7
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    Hello, PythagorasNeophyte!

    $\displaystyle 1.\;\;\dfrac{(a-b)^2 +a(a+b) -11b^2}{a^2 -4b^2}$

    $\displaystyle \text{Here are my workings:}$

    $\displaystyle \dfrac{(a-b)^2 +a(a+b) -11b^2}{a^2 -4b^2}$

    . . $\displaystyle =\; \dfrac{a^2-2ab+b^2+a^2+ab-11b^2}{(a-2b)(a+2b)}$

    . . $\displaystyle =\;\dfrac{a^2+a^2-2ab+ab+b^2-11b^2}{(a-2b)(a+2b)}$

    . . $\displaystyle =\;\dfrac{2a^2-ab-10b^2}{(a-2b)(a+2b)}$

    . . $\displaystyle =\;\dfrac{(-2a+5b)(-a-2b)}{(a-2b)(a+2b)}$ .
    ?

    Why did you factor like that?



    If I asked you do reduce $\displaystyle \frac{6}{15}$, would you do this?

    . . . . . . $\displaystyle \dfrac{6}{15} \;=\;\dfrac{(\text{-}2)(\text{-}3)}{(3)(5)}$

    . . then say "Now what do I do?"

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