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Equation Involving Complex Fraction

  1. May 19, 2014 #1
    1. The problem statement, all variables and given/known data

    Solve the formula f=1/1/a+1/b for a in terms of b and f.



    3. The attempt at a solution

    My attempt at this problem was to first multiply all terms by ab, thus getting rid of the fractions in the denominator on the right. I was then left with the equation fab=ab/(b+a), so I multiplied each side by (b+a), the equation now looks like (b+a)fab=ab, I then divided by ab, and was left with f(b+a)=1, I then dived by f , and subtract b, and am left with; a=1-b/f. Now, the correct answer, according to my textbook, is a =bf/b-f. All I can find on the internet or in my textbook is how to deal with a an expression containing complex fractions, but nothing on equations containing complex fractions, so I'm at somewhat of a loss as to what I'm doing wrong. Anyway, I appreciate any guidance on how to solve this that anyone has to offer.


    Also, an apology for not using Latex. I'm on
    my phone, and it looked as if it doesn't work on phones. I tried my best to be as clear as possible.
     
  2. jcsd
  3. May 19, 2014 #2

    tms

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    Use parentheses to clarify your equation, if you can't use LaTeX. To me it looks like
    $$a + \frac{1}{b}.$$
    or maybe
    $$\frac{1}{a} + \frac{1}{b}.$$
     
  4. May 19, 2014 #3
    Sorry, let me retry that: f=1/[(1/a)+(1/b)]. I hope that makes a little more sense.
     
  5. May 19, 2014 #4

    haruspex

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    You should not multiply "all terms" by ab.
    You can multiply each side of the equality by ab; or you can multiply the numerator and denominator of the fraction by ab.
     
  6. May 19, 2014 #5

    I described that incorrectly. I meant that I had multiplied the numerators.
     
  7. May 19, 2014 #6

    haruspex

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    In order to get
    you would have done what you said you did. You don't get that just by multiplying the numerators each side, or (more usefully) just by multiplying the numerator and denominator on the RHS.
     
  8. May 19, 2014 #7
    Ok. This may sound stupid, but I was under the impression one could not just multiply the right hand side(what I assume RHS means). I can get on a computer now and I'm going to try using LaTex.
     
  9. May 19, 2014 #8
    [tex]f= \frac{1}{1/a + 1/b}[/tex]

    [tex]ab * f = \frac{1}{1/a + 1/b} * ab[/tex]

    [tex]fab = \frac{ab}{b+a}[/tex]

    [tex]fab{(b+a)} = ab[/tex]

    [tex] \frac{f(b+a)ab}{ab} = \frac{ab}{ab} = f{(b + a)} = 1[/tex]

    [tex] a = \frac{1}{f} - b[/tex]

    That's what I came up with. It looks horribly wrong even as I type it, yet, for some I reason can't quite think of another way to do this.
     
    Last edited: May 19, 2014
  10. May 19, 2014 #9
    Is ##\frac{1}{2}*3=\frac{1*3}{2*3}=\frac{3}{6}=\frac{1}{2}##, or is ##\frac{1}{2}*3=\frac{1*3}{2}=\frac{3}{2}##?
     
  11. May 19, 2014 #10
    This is where I was getting confused. I'm aware that 3=3/1, I just haven't seen how I could multiply away the denominator of a denominator. Does this look right as the next step, as it seems like the only other logical way I can think of.

    [tex]\frac{1}{ab} • f = \frac{1}{1/a + 1/b} • \frac{1}{ab} = \frac{f}{ab} = \frac{1}{b + a}[/tex]
     
    Last edited: May 19, 2014
  12. May 19, 2014 #11
    [tex] \frac{1}{ab} • f = \frac{1}{1/a + 1/b} • \frac{1}{ab} = ab = fb + fa = ab - fa= a{(b-f)} =fb =a = \frac{fb}{b-f}[/tex]

    I feel like a total dunce. I used the advice given, which should have been obvious to me, and multiplied by 1/ab, which of course gave the correct answer. Thanks for the suggestions everyone.
     
  13. May 19, 2014 #12

    haruspex

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    This is the faulty step. [tex]\frac{1}{1/a + 1/b} * ab = \frac{ab}{1/a + 1/b}[/tex]
     
  14. May 19, 2014 #13

    Yeah, I didn't make sense of it until I really thought about gopher's post, than I realized that

    [tex] \frac{1}{ab}[/tex]

    Is the same as

    [tex]\frac{1}{ab/1}[/tex]

    Which finally drilled into my brain the proper way to solve the equation for a. Such an obvious solution that somehow eluded me...
     
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