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Math Help - Can somebody please show me how to solve these 2 problems step by step?

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    Post Can somebody please show me how to solve these 2 problems step by step?

    5. [12x/ (5xy^4)] - [9y^2/(15x^3y^2)]

    6. [6/(x+2)] + 11/ (x-7)
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    6. [6/(x+2)] + 11/ (x-7)
    We find the LCD of the two fractions and that is (x+2)(x-7)

    So now we build the fraction to had the LCD

    \frac{6}{(x+2)}\frac{(x-7)}{(x-7)}+\frac{11}{(x-7)}\frac{(x+2)}{(x+2)}=\frac{6x-42}{ (x-7)(x+2) }+\frac{11x+22}{(x-7)(x+2)}

    =\frac{17x-20}{(x-7)(x+2)}
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    #5 Answer

    [12x/ (5xy^4)] - [9y^2/(15x^3y^2)]

    LCD = 15 x^3 y^4

    =[12x (3 x ^2) / 15 x^3 y^4] - [9y^2 (y^2) / 15 x^3 y ^4]

    =[36 x^3 / 15x^3y^4] - [9y^4 / 15x^3y^4]

    = 36x^3 - 9y^4 / 15x^3y^4
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    Quote Originally Posted by AlgebraAnswers View Post
    [12x/ (5xy^4)] - [9y^2/(15x^3y^2)]

    LCD = 15 x^3 y^4

    =[12x (3 x ^2) / 15 x^3 y^4] - [9y^2 (y^2) / 15 x^3 y ^4]

    =[36 x^3 / 15x^3y^4] - [9y^4 / 15x^3y^4]

    = 36x^3 - 9y^4 / 15x^3y^4
    This can be simplified to

    \frac{36x^3-9y^4}{15x^3y^4}=\frac{9(4x^3-y^4)}{15x^3y^4}=\frac{3(4x^3-y^4)}{5x^3y^4}
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