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Simple algebra question

  1. Sep 19, 2005 #1
    I am trying to figure out an overall reaction constant, but I am getting stuck on the math behind the problem. I am at this point in the problem:

    [Co(H2O)6] = [Co] x (k[H+]/(1+k[H+]))
    And I am trying to mathematically get to
    [Co(H2O)6] = [Co] x (1-(1/(1+[H+]k)))

    When you ignore the chemistry, the equation is the same as:

    a = b x (cd/(1+cd))

    going to

    a = b x (1-(1/(1+cd)))

    How do you factor out the cd from the numerator?

  2. jcsd
  3. Sep 19, 2005 #2


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    It wasn't factored out. (If it was factored out, it would still be there, but outside. :tongue2:)

    Try going the other direction: what sort of things can you do to the latter equation?
  4. Sep 19, 2005 #3
    So if I work backwards, the step before 1-(1/(1+cd)) could have possibly been
    [(1+cd)/(1+cd)] - (1/(1+cd)). What would have led me to that step or is that not the correct step backwards?
  5. Sep 19, 2005 #4
    Wait a minute! I think a lightbulb just went off...
  6. Sep 19, 2005 #5
    [(1+cd)/(1+cd)] - (1/(1+cd)) This actually reduces back to cd/(1+cd), right? Because the 1-1=0 and cd is the only thing left in the numerator. So what would prompt me to know that I should change cd/(1+cd) into [(1+cd)/(1+cd)] - (1/(1+cd))?
  7. Sep 19, 2005 #6


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    Well, whether you should change it is another question entirely.

    But to figure out how to change it, you can simply use the division algorithm.

    You're dividing cd by (cd + 1)...

    How many times does (cd + 1) go into cd? 1 time, with remainder -1.

    So, cd/(cd+1) = 1 + (-1)/(cd+1)

    (when figuring out how many times, you just look at the "biggest" terms -- in this case, we consider any term involving "cd" to be bigger than any term that is just a number)

    Another method is to look at it and think "Hrm, I can probably write that as A + B/(cd+1)", and then try to solve the equation "A + B/(cd+1) = cd/(cd+1)"

    A third way is to learn how to do what you just did in the opposite direction!
    Last edited: Sep 19, 2005
  8. Sep 19, 2005 #7
    Thanks so much!
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