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Math Help - LCD help

  1. #1
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    LCD help

    Could someone please explain to me how they get this LCD?



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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, silencecloak!

    That is the dumbest way to write the problem ... and the answer.


    Could someone please explain to me how they get this LCD?

    -\frac{2^2}{(-x+3)(x+3)} - \frac{2x+1}{x(x-3)} \;=\;\frac{x(-2^2) + [-(x+3)][-(2x+1)]}{(-x+3)(x+3)x}

    I'd write it as: . -\frac{4}{-(x-3)(x+3)} - \frac{2x+1}{x(x-3)} \;=\; \frac{4}{(x-3)(x+3)} - \frac{2x+1}{x(x+3)}


    The LCD is x(x-3)(x+3): . {\color{blue}\frac{x}{x}}\cdot\frac{4}{(x-3)(x+3)} \:- \:{\color{blue}\frac{x+3}{x+3}}\cdot\frac{2x+1}{x(  x-3)}

    . . = \;\;\frac{4x}{x(x-3)(x+3)} - \frac{(x+3)(2x+1)}{x(x-3)(x+3)} \;\;=\;\;\frac{4x - (x+3)(2x+1)}{x(x-3)(x+3)}\;\;\hdots\;\;\text{etc.}

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