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Math Help - subtracting rationals

  1. #1
    Member Jonboy's Avatar
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    subtracting rationals

    Hi Everyone! I would like you to check my work.

    BTW my book is Precalculus, 3rd Edition by Bittinger, Beecher, Ellenbogen, and Penna

    The directions are to add/subtract/divide and simplify if possible.

    #38: \frac{3a}{3a\,-\,2b}\,-\,\frac{2a}{2b\,-\,3a}

    Get like denominators: \;\frac{3a}{3a\,-\,2b}\,-\,\frac{2a}{-3a\,+\,2b}

    \;=\frac{-3a}{-3a\,+\,2b}\,-\,\frac{2a}{-3a\,+\,2b}

    Simplify: \;\frac{\,-\,5a}{-3a\,+\,2b}

    Would it be better to put my answer as: \;\frac{5a}{3a\,-\,2b} ?

    # 39: \frac{9x\,+\,2}{3x^2\,-\,2x\,-\,8}\,+\,\frac{7}{3x^2\,+\,x\,-\,4}

    Factor the denominators: \frac{9x\,+\,2}{(3x\,+\,4)(x\,-\,2)}\,+\,\frac{7}{(3x\,+\,4)(x\,-\,1)}

    Get a LCM of the denominators: \frac{(9x\,+\,2)(x\,-\,1)}{(3x\,+\,4)(x\,-\,2)(x\,-\,1)}\,+\,\frac{7x\,-\,14}{(3x\,+\,4)(x\,-\,1)(x\,+\,2)}

    \;=\,\frac{9x^2\,-\,7x\,-\,2\,+\,7x\,-\,14}{(3x\,+\,4)(x\,-\,2)(x\,-\,1)}

    Simplify: \frac{(3x\,-\,4)(3x\,+\,4)}{(3x\,+\,4)(x\,-\,2)(x\,-\,1)}

    Which simplifies to:

    \frac{3x\,-\,4}{(x\,-\,2)(x\,-\,1)}

    My book agrees so I'm pretty sure.

    #44: \frac{6}{x\,+\,3}\,-\,\frac{x\,+\,4}{9\,-\,x^2}\,+\,\frac{2x\,-\,3}{9\,-\,6x\,+\,x^2}

    My answer is: \frac{-8x^2\,+\,32x\,-\,41}{-(x\,-\,3)^2(x\,+\,3)}

    # 62: \frac{1}{x^2}\,-\,\frac{1}{y^2}\div\frac{1}{x^2}\,-\,\frac{2}{xy}\,+\,\frac{1}{y^2}

    Get the like denominators: \frac{y^2\,-\,x^2}{x^2y^2}\,\div\,\frac{y^2\,-\,2x\,+\,x^2}{x^2y^2}

    \;=\,\frac{y^2\,-\,x^2}{x^2y^2}\cdot\,\frac{x^2y^2}{y^2\,-\,2x\,+\,x^2}

    Which just simplifies to: \frac{y^2\,-\,x^2}{y^2\,-\,2x\,+\,x^2}

    Correct?

    Thank you all for your help!
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  2. #2
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      \frac{(3x-4)(3x+4)}{(3x+4)(x-2)(x-1)} = \frac{3x-4}{(x-1)(x-2)}
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  3. #3
    Member Jonboy's Avatar
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    Quote Originally Posted by shilz222 View Post
      \frac{(3x-4)(3x+4)}{(3x+4)(x-2)(x-1)} = \frac{3x-4}{(x-1)(x-2)}
    yeah i just saw that; thanks
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jonboy View Post
    Hi Everyone! I would like you to check my work.

    BTW my book is Precalculus, 3rd Edition by Bittinger, Beecher, Ellenbogen, and Penna

    The directions are to add/subtract/divide and simplify if possible.

    #38: \frac{3a}{3a\,-\,2b}\,-\,\frac{2a}{2b\,-\,3a}

    Get like denominators: \;\frac{3a}{3a\,-\,2b}\,-\,\frac{2a}{-3a\,+\,2b}

    \;=\frac{-3a}{-3a\,+\,2b}\,-\,\frac{2a}{-3a\,+\,2b}

    Simplify: \;\frac{\,-\,5a}{-3a\,+\,2b}
    Good!

    Would it be better to put my answer as: \;\frac{5a}{3a\,-\,2b} ?
    yes. it looks nicer, which is good

    # 39: \frac{9x\,+\,2}{3x^2\,-\,2x\,-\,8}\,+\,\frac{7}{3x^2\,+\,x\,-\,4}

    Factor the denominators: \frac{9x\,+\,2}{(3x\,+\,4)(x\,-\,2)}\,+\,\frac{7}{(3x\,+\,4)(x\,-\,1)}

    Get a LCM of the denominators: \frac{(9x\,+\,2)(x\,-\,1)}{(3x\,+\,4)(x\,-\,2)(x\,-\,1)}\,+\,\frac{7x\,-\,14}{(3x\,+\,4)(x\,-\,1)(x\,+\,2)}

    \;=\,\frac{9x^2\,-\,7x\,-\,2\,+\,7x\,-\,14}{(3x\,+\,4)(x\,-\,2)(x\,-\,1)}

    Simplify: \frac{(3x\,-\,4)(3x\,+\,4)}{(3x\,+\,4)(x\,-\,2)(x\,-\,1)}

    Which simplifies to:

    \frac{3x\,-\,4}{(x\,-\,2)(x\,-\,1)}

    My book agrees so I'm pretty sure.
    yes, good. i was just about to tell you it was the difference of two squares, but you beat me to it. nice

    By the way, you LaTex is way more complicated than it has to be. there is no need to put "\," after every term
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Jonboy View Post
    #44: \frac{6}{x\,+\,3}\,-\,\frac{x\,+\,4}{9\,-\,x^2}\,+\,\frac{2x\,-\,3}{9\,-\,6x\,+\,x^2}

    My answer is: \frac{-8x^2\,+\,32x\,-\,41}{-(x\,-\,3)^2(x\,+\,3)}
    my answer is \frac {9x^2 - 32x + 33}{(x + 3)(x - 3)^2}


    # 62: \frac{1}{x^2}\,-\,\frac{1}{y^2}\div\frac{1}{x^2}\,-\,\frac{2}{xy}\,+\,\frac{1}{y^2}

    Get the like denominators: \frac{y^2\,-\,x^2}{x^2y^2}\,\div\,\frac{y^2\,-\,2x\,+\,x^2}{x^2y^2}

    \;=\,\frac{y^2\,-\,x^2}{x^2y^2}\cdot\,\frac{x^2y^2}{y^2\,-\,2x\,+\,x^2}

    Which just simplifies to: \frac{y^2\,-\,x^2}{y^2\,-\,2x\,+\,x^2}
    my answer is: \frac {(x - y)^2 - x^4}{x^2 y^2} = \frac {\left( x - y + x^2 \right) \left( x - y - x^2 \right)}{x^2 y^2}

    remember the order of operations. we do division before addition and subtraction


    the only way your answer would be valid is if the question originally said \left( \frac{1}{x^2}\,-\,\frac{1}{y^2} \right) \div \left( \frac{1}{x^2}\,-\,\frac{2}{xy}\,+\,\frac{1}{y^2} \right)
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