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Math Help - Simplifying Algebraic expressions & Adding fractions with unlike denominators

  1. #1
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    Simplifying Algebraic expressions & Adding fractions with unlike denominators

    I have two questions, both of which I am stuck on:

    Problem #1:

    Instructions state, 'Simplify the expression:'

    y - x / (x^2 + xy - 2y^2)

    Problem #2:

    Instructions state, 'Perform the indicated operation and simplify if possible:'

    3 / (x^2 - 9) + 2 / (x - 3)

    I've determined that x^2 - 9 can be factored to (x + 3) (x - 3) but I don't know where to take it from here. Do I have to find an LCD to solve this?
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  2. #2
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    Re: Simplifying Algebraic expressions & Adding fractions with unlike denominators

    Hello, Odonsky!


    \text{1. Simplify: }\:\frac{y - x}{x^2 + xy - 2y^2}

    \text{Factor: }\:\frac{-(x-y)}{(x-y)(x+2y)}

    \text{Reduce: }\:\frac{-({\color{red}\rlap{/////}}x-y)}{({\color{red}\rlap{/////}}x-y)(x+2y)} \;=\;\frac{\text{-}1}{x+2y}



    \text{2. Simplify: }\:\frac{3}{x^2 - 9} + \frac{2}{x - 3}

    \text{Factor: }\:\frac{3}{(x-3)(x+3)} + \frac{2}{x-3} \;=\;\frac{3}{(x-3)(x+3)} + \frac{2}{x-3}\cdot\frac{x+3}{x+3}

    . . . . . =\;\frac{3}{(x-3)(x+3)} + \frac{2(x+3)}{(x-3)(x+3)} \;=\;\frac{3 + 2(x+3)}{(x-3)(x+3)}

    . . . . . =\;\frac{3+2x+6}{(x-3)(x+3)} \;=\;\frac{2x+9}{(x-3)(x+3)}
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  3. #3
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    Re: Simplifying Algebraic expressions & Adding fractions with unlike denominators

    You are right find LCD of denominator which is ( x+3)(x-3) now we combine the fractions and get [ 3 + 2 (x+3)]/(x+3)(x-3) = (2x +9)/(x^2-9)
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  4. #4
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    Re: Simplifying Algebraic expressions & Adding fractions with unlike denominators

    Quote Originally Posted by Soroban View Post
    Hello, Odonsky!



    \text{Factor: }\:\frac{-(x-y)}{(x-y)(x+2y)}

    \text{Reduce: }\:\frac{-({\color{red}\rlap{/////}}x-y)}{({\color{red}\rlap{/////}}x-y)(x+2y)} \;=\;\frac{\text{-}1}{x+2y}
    Thank you for the detailed response!

    However, I do not quite understand why you factored a negative 1 into the numerator. Would you please care to explain?

    Thank you.
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  5. #5
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    Re: Simplifying Algebraic expressions & Adding fractions with unlike denominators

    Odonsky has correctly done what he has. he has just taken out (-1) common so that he gets the same factor in numerator i.e., ( x-y) which is also there in the denominator for cancellation.
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