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Math Help - Addition and subtraction with fractions Factorization

  1. #1
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    Addition and subtraction with fractions Factorization

    I used Google Translate for the title because my English isn't that good...

    Anyway,
    What am I doing wrong?
    Here's my attempt to solve this exercise:

    (http://img52.imageshack.us/img52/1108/question.png)

    And here's the solution:

    (http://img21.imageshack.us/img21/6429/answerz.png)

    Can anyone explain me how to solve that right?
    Thanks!
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  2. #2
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    your mistake is in the numerator of the 3rd expression ...

    \frac{(a-b)(a+b) + 3ab + b^2 - 4ab}{3b(a-b)}

    \frac{a^2 - b^{\textcolor{red}{2}} + 3ab + b^2 - 4ab}{3b(a-b)}

    \frac{a^2 - ab}{3b(a-b)}

    \frac{a(a - b)}{3b(a-b)}

    \frac{a}{3b}
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  3. #3
    MHF Contributor red_dog's Avatar
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    You did a mistake: (a-b)(a+b)=a^2-b^2 not a^2-b.
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  4. #4
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    Thanks!!!
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  5. #5
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    The first couple of steps are correct. You have
    \frac{a+b}{3b}+ \frac{a}{a-b}+ \frac{b-4a}{3a-3b} and correctly recognize that the "3" in that last denominator can be factored out:
    \frac{a+b}{3b}+ \frac{a}{a-b}+ \frac{b-4a}{3(a-b)}
    so the common denominator is 3b(a-b).

    Multiplying both numerator and denominator of: the first fraction by (a-b); the second fraction by 3b; the third fraction by b gives

    \frac{(a+b)(a-b)+ a(3b)+ (b- 4a)(b)}{3b(a-b)}

    But now you have a^2- b as the first two terms in the next fraction and (a+b)(a-b)= a^2- b^2. That was probably a typo but you continued it through your answer. You should have
    \frac{a^2- b^2+ 3ab+ b^2- 4ab}{3b(a-b)}.

    Blast! Skeeter beat me to it again!

    Now the " b^2" terms cancel and the fraction simplifies.
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