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Math Help - Rational Expression Help

  1. #1
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    Rational Expression Help

    Im having trouble with a problem off a review... Ill try to my best abilities to write it out


    Solve.



    3 OVER x^2 - 6x + 9

    PLUS

    x - 2 OVER 3x - 9


    SET EQUAL TO

    x OVER 2x - 6


    First I factor out all the denominators. Get the LCD then multiply it to each individual term to get rid of my fractions..

    But this is where I have a problem. I dont understand what the LCD should be. Should it be 6(x-3)(x-3)?
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  2. #2
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    Is this the equation?

    \displaystyle \frac{3}{x^2-6x+9}+\frac{x-2}{3x-9}= \frac{2}{2x-6}
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  3. #3
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    So you are trying to solve for \displaystyle x in the equation

    \displaystyle \frac{3}{x^2 - 6x + 9} + \frac{x-2}{3x - 9} = \frac{x}{2x-6}.

    For starters, factor all the denominators

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

    Now get a common denominator for both sides, and yes, you are correct that the lowest common denominator is \displaystyle 6(x - 3)^2.
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  4. #4
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    ok.. I multiply 6(x-3)^2 to each section and I get..


    18 + 2x-6 = 3x -9

    then i try and solve for x and get 21... which isnt the write answer according to this answer sheet
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  5. #5
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    Quote Originally Posted by gurrry View Post
    ok.. I multiply 6(x-3)^2 to each section and I get..


    18 + 2x-6 = 3x -9

    then i try and solve for x and get 21... which isnt the write answer according to this answer sheet
    OK, for starters, don't jump to multiplying by the common denominator. I ALWAYS write the fractions with their common denominators first, it is clearer and reduces mistakes.

    So \displaystyle \frac{6\cdot 3}{6(x - 3)^2} + \frac{2(x - 3)(x - 2)}{6(x - 3)^2} = \frac{3x(x - 3)}{6(x - 3)^2}

    \displaystyle \frac{18}{6(x - 3)^2} + \frac{2x^2 - 10x + 12}{6(x - 3)^2} = \frac{3x^2 - 9x}{6(x- 3)^2}

    \displaystyle \frac{2x^2 - 10x + 30}{6(x - 3)^2} = \frac{3x^2 - 9x}{6(x - 3)^2}

    \displaystyle 2x^2 - 10x + 30 = 3x^2 - 9x.

    Now solve the resulting quadratic.
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  6. #6
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    Quote Originally Posted by Prove It View Post
    OK, for starters, don't jump to multiplying by the common denominator. I ALWAYS write the fractions with their common denominators first, it is clearer and reduces mistakes.

    So \displaystyle \frac{6\cdot 3}{6(x - 3)^2} + \frac{2(x - 3)(x - 2)}{6(x - 3)^2} = \frac{3x(x - 3)}{6(x - 3)^2}

    \displaystyle \frac{18}{6(x - 3)^2} + \frac{2x^2 - 10x + 12}{6(x - 3)^2} = \frac{3x^2 - 9x}{6(x- 3)^2}

    \displaystyle \frac{2x^2 - 10x + 30}{6(x - 3)^2} = \frac{3x^2 - 9x}{6(x - 3)^2}

    \displaystyle 2x^2 - 10x + 30 = 3x^2 - 9x.

    Now solve the resulting quadratic.
    So for the 2(x-3)(x-2) part, did you just multiply out (x-3)(x-2) then go back and multiply 2 throughout the entire thing? Or do you distribute the 2 through the x-3 first then use the FOIL method?
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  7. #7
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    Quote Originally Posted by gurrry View Post
    So for the 2(x-3)(x-2) part, did you just multiply out (x-3)(x-2) then go back and multiply 2 throughout the entire thing? Or do you distribute the 2 through the x-3 first then use the FOIL method?
    It doesn't make any difference, multiplication is associative...
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