Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Transform this equatio

  1. Feb 12, 2007 #1
    1. The problem statement, all variables and given/known data

    Need to rearrange this equation to make b subject
    [tex]R = \frac{ab}{a+b}[/tex]

    2. The attempt at a solution

    R(a+b) = ab
    Ra +Rb = ab
    [tex]\frac{Rb}{b} = a [/tex] x -Ra

    i think i did the last step wrong. Where do i go to now?

  2. jcsd
  3. Feb 12, 2007 #2
    after Ra + Rb = ab

    using addition and subtraction you need to get all terms that have a factor b on one side of the equation and all terms that do not have a factor of b on the other side of the equation
  4. Feb 12, 2007 #3
    ive tried

    [tex]\frac{Rb}{b} = a x -Ra[/tex]

    but i think that's wrong

    can you show me what you'd do here?
  5. Feb 12, 2007 #4
    Yes it wrong. Go back to the step where you were correct (see my last post) and do not do any division at this stage. Have another read of my post and let me know if you do not understand what I am suggesting you do.
  6. Feb 12, 2007 #5

    Ra + Rb = ab

    can be written as

    Ra + Rb = a x b

    dont you cancel out x b with a /b (divide b)

    If not, and you just use +/- then isn't it

    Ra + Rb = ab
    Rb - b = a - Ra
    b(R-1) = a - Ra
    b = a - Ra/(R-1)
    But that doesn't look right to me...
  7. Feb 12, 2007 #6


    User Avatar
    Staff Emeritus
    Science Advisor

    No, this is not right. You have: Ra+Rb=ab. Collect the terms including b together on one side: Ra=ab-Rb.

    Now, can you factorise the right hand side? Once you have it factorised, it should be easy to make b the subject.
  8. Feb 12, 2007 #7
    No you cannot separate the ab except by division

    Ra + Rb = ab subtract Rb from both sides
    Ra = ab -Rb factorise
    Ra = b(a - R) now do the division to obtain b
  9. Feb 12, 2007 #8
    Ra = ab - Rb
    b(a-R) = Ra
    b = Ra/(a - R)
  10. Feb 12, 2007 #9
    Yes it is right
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook