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Need some guidance - Fractions

  1. Aug 3, 2013 #1
    1. The problem statement, all variables and given/known data

    Solve for b. ##\frac{4}{3b-2}-\frac{7}{3b+2}=\frac{1}{9b^2-4}##


    3. The attempt at a solution

    ##\frac{4}{3b-2}-\frac{7}{3b+2}=\frac{1}{(2b-2)(3b+2)}##

    ##(3b-2)(3b+2)\left(\frac{4}{3b-2}\right)-\left(\frac{7}{3b+2}\right)(3b-2)(3b+2)=\left(\frac{1}{(3b-2)(3b+2)}\right)(3b-2)(3b+2)##

    After cancelling this messy expression I get: ##\frac{12b+8}{(3b-2)(3b+2)}-\frac{21b-14}{(3b-2)(3b+2)}=1##

    Still pretty messy so I will expand and cancel further ##\frac{4}{3b-2}-\frac{7}{3b+2}=1##

    Now I am stuck :/ I'm not sure if this is even correct but I dunno how to proceed.

    **EDIT** I think I see what I did wrong.

    I can just cross multiply right from the beginning giving me ##4(3b-2)-7(3b+2)=\frac{1}{(3b-2)(3b+2)}## but not sure what to do now... lol I'm sure this is the sort of thing I should be doing.
     
    Last edited: Aug 3, 2013
  2. jcsd
  3. Aug 3, 2013 #2

    tiny-tim

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    Hi MathJakob! :smile:
    Yup! :biggrin:

    except that RHS should just be 1, shouldn't it? :wink:
     
  4. Aug 3, 2013 #3
    Why should it just be 1?
     
  5. Aug 3, 2013 #4

    tiny-tim

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    because you're multiplying both sides by (3b+2)(3b-2),

    and the RHS is only 1/(3b+2)(3b-2) to start with! :wink:

    EDIT: ooh, just noticed:

    your 4(3b−2)−7(3b+2) should actually have been 4(3b+2)−7(3b-2)
     
  6. Aug 3, 2013 #5
    oh yeh silly me D: this always happens to me when I'm doing fractions. After a while I forget even the most basic of rules... sometimes I even forget that ##\frac{x}{1}## is really just ##x## fractions are actually one of the more difficult aspects of mathematics for me...

    Sometimes I just go completely brain dead, really annoying :P
     
  7. Aug 4, 2013 #6

    Curious3141

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    I didn't really read the preceding, but this is definitely not right. You've completely omitted the common denominator on the LHS.

    When you combine the expressions on the LHS (cross-multiply), the denominator becomes ##(3b+2)(3b-2) = 9b^2 - 4##.

    There's also a sign mixup in what you wrote. So what you should end up with is:

    ##\displaystyle \frac{4(3b+2) - 7(3b-2)}{9b^2-4} = \frac{1}{9b^2 - 4}##

    Solving that is easy, just multiply both sides by ##(9b^2 - 4)##. An important caveat here is that you cannot multiply both sides of an equation by zero and hope to get valid solutions. So ##9b^2 - 4 \neq 0 \implies b \neq \pm\frac{2}{3}##, so you should exclude those solutions if you happen to encounter them when you work through the algebra. In this case, you won't see them.
     
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