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Math Help - Complex fraction simplification

  1. #1
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    Smile Complex fraction simplification

    Good afternoon forum I am hav[IMG]file:///C:/DOCUME%7E1/Owner/LOCALS%7E1/Temp/moz-screenshot-3.jpg[/IMG]ing problems with the below problem


    [IMG]file:///C:/DOCUME%7E1/Owner/LOCALS%7E1/Temp/moz-screenshot-4.jpg[/IMG]
    Attached Thumbnails Attached Thumbnails Complex fraction simplification-p11-p.5-56-p.-63.jpg  
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  2. #2
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    Hello, AlgebraicallyChallenged!

    Simplify: . \frac{\dfrac{x+2}{x^2-1} + \dfrac{1}{x+1}} {\dfrac{x}{2x^2-x-1} + \dfrac{1}{x-1}}
    Factor all denominators: . \frac{\dfrac{x+2}{(x-1)(x+1)} + \dfrac{1}{x+1}} {\dfrac{x}{(x-1)(2x+1)} + \dfrac{1}{x-1}}

    The LCD is (x-1)(x+1)(2x+1)


    Multiply top and bottom by the LCD:

    . . \frac{(x-1)(x+1)(2x+1)\left[\dfrac{x+2}{(x-1)(x+1)} + \dfrac{1}{x+1}\right]} {(x-1)(x+1)(2x+1)\left[\dfrac{x}{(x-1)(2x+1)} + \dfrac{1}{x-1}\right]} . = \;\frac{(2x+1)(x+2) + (x-1)(2x+1)}{x(x+1) + (x+1)(2x+1)}

    . . = \;\frac{2x^2 + 5x + 2 + 2x^2 - x - 1}{x^2+x + 2x^2+3x+1} \;=\;\frac{4x^2 + 4x + 1}{3x^2+4x+1} \;=\;\boxed{\frac{(2x+1)^2}{(x+1)(3x+1)}} . . . doesn't reduce

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  3. #3
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    Quote Originally Posted by AlgebraicallyChallenged View Post
    Good afternoon forum I am hav[IMG]file:///C:/DOCUME%7E1/Owner/LOCALS%7E1/Temp/moz-screenshot-3.jpg[/IMG]ing problems with the below problem


    [IMG]file:///C:/DOCUME%7E1/Owner/LOCALS%7E1/Temp/moz-screenshot-4.jpg[/IMG]
     \frac{ \frac{x+2}{x^2+1}+\frac{1}{x+1} }{\frac{x}{2x^2-x-1}+\frac{1}{x-1}}


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


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


     = \frac{(x+2)(x+1)+x^2+1}{(x^2+1)(x+1)} \times \frac{(2x^2-x-1)(x-1)}{x(x-1) + 2x^2-x-1}

    How's that?

    [Beaten to it]
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