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Math Help - Addition and substraction of algebraic fractions -urgent questions

  1. #1
    Junior Member PythagorasNeophyte's Avatar
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    Addition and substraction of algebraic fractions -urgent questions

    A very good day to everyone here on Math Help Forum!
    I have a few questions related to algebraic fractions. They are as the following:

    1. \dfrac{x^2}{y} + \dfrac{y}{x^2}

    2. \dfrac{x}{x+16} +  \dfrac{5}{x^2 +x-12}

    3. \dfrac{5}{2x^2 +5x +3}  - \dfrac{1}{2x^2 +3x -2}

    My workings are shown below:

    1. \dfrac{x^2}{y} + \dfrac{y}{x^2}

     = \dfrac{x^2y+x^2y}{x^2y}

    =\dfrac{2x^2y}{x^2y}

     <br />
=2<br />
?

    2. \dfrac{x}{x+16} +  \dfrac{5}{x^2 +x -12}

    =\dfrac{x}{x^2 -4^2} + \dfrac{5}{(x+4)(x-3)}

    =\dfrac{x}{(x-4)(x+4)} + \dfrac{5}{(x+4)(x-3)}

     <br />
=\dfrac{x(x-3)+5(x-4)}{(x-4)(x+4)(x-3)}<br />


     <br />
=\dfrac{x^2+2x-20}{(x-4)(x+4)(x-3)}<br />

     <br />
 =\dfrac{(x+6)(x-4)}{(x-4)(x+4)(x-3)}<br />

     <br />
= \dfrac{x+6}{(x+4)(x-3)}<br />
?



    3. \dfrac{5}{2x^2 +5x +3}  - \dfrac{1}{2x^2 +3x -2}

    =\dfrac{5}{(2x+3)(x+1)} - \dfrac{1}{(2x-1)(x+2)}

     <br />
=\dfrac{5(2x-1)(x+2) -1(2x+3)(x+2)}{(2x+3)(x+1)(2x-1)(x+2)}<br />

    This is where I am stuck.

    If you happened to notice any mistakes and blunders, please pinpoint them and explain how it should be done.
    Thank you so much for taking your time!
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  2. #2
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    Quote Originally Posted by PythagorasNeophyte View Post
    A very good day to everyone here on Math Help Forum!
    I have a few questions related to algebraic fractions. They are as the following:

    1. \dfrac{x^2}{y} + \dfrac{y}{x^2}
    Incorrect. To get a common denominator, multiply the top and bottom of each fraction by the same thing (i.e. a cleverly disguised 1)...

    \frac{x^2}{y} + \frac{y}{x^2} = \frac{x^2}{y}\left(\frac{x^2}{x^2}\right) + \frac{y}{x^2}\left(\frac{y}{y}\right)

     = \frac{x^4}{x^2y} + \frac{y^2}{x^2y}

     = \frac{x^4 + y^2}{x^2y}.
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  3. #3
    Senior Member Educated's Avatar
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    2. \dfrac{x}{x+16} + \dfrac{5}{x^2 +x -12} = \dfrac{x}{x+16} + \dfrac{5}{(x+4)(x-3)}

    You will then need to find a cross multiply to get a common denominator.

    Note: \dfrac{x}{x+16} \ne \dfrac{x}{x^2 - 4^2}


    3. Expand the numerator, gather up like terms and then factorise again.
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  4. #4
    Junior Member PythagorasNeophyte's Avatar
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    Hello Educated,

    You said that \dfrac{x}{x+16} + \dfrac{5}{x^2 +x -12} = \dfrac{x}{x+16} + \dfrac{5}{(x+4)(x-3)} needs to be cross multiplied. But it isn't an equation. It's about the addition of algebriac fractions.

    I wonder if x+16 can be factorised into x^2+4^2 then into (x+4)(x+4)<br />
. If it's possible, then the LCM would be (x+4)(x+4)(x-3).

    Otherwise, the LCM would be (x+16)(x+4)(x-3).

    Which of these answers could be correct?
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  5. #5
    Senior Member Educated's Avatar
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    LCM is (x+16)(x+4)(x-3)

    No it is not an equation, but you can still cross multiply the equation to get a common denominator to simplify it:

    \dfrac{x(x+4)(x-3)}{(x+16)(x+4)(x-3)} + \dfrac{5(x+16)}{(x+16)(x+4)(x-3)}

    I was just trying to state that x + 16 doesn't equal x^2 + 4^2. You cannot factorise x + 16.
    Last edited by Educated; September 20th 2010 at 02:43 AM.
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