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Transposing formulas/Solving for variables

  1. Mar 4, 2013 #1

    I have a question about solving or isolating for variables in an equation. It doesnt need to be an equation, however, I've noticed my problems always appear when trying to transpose them. I have read through algebra concepts, but they only tell me what I already know regarding polynomial rules and such. I think my problem is much more basic:

    Keq = [CO2][CF4] / [COF2]2

    CO2 = [COF2]2 / [CF4]

    What exactly happens that causes the top and bottom parts to reverse their orientation? What caused the CO2 to appear on the left side, what operation/method is used?

    Are there any posts similar to this that have a detailed explanation? Reading a book about Algebra didnt help the second time around, so I would just like to know how the rules are applied. What if there are two sets of variables in the denominator, can one of them be moved without altering the other? When a square root contains a variable of interest and is being multiplied by another variable in the denominator, what can be done to isolate it? It's these kind of things I've had problems with for a while. Could someone explain?
  2. jcsd
  3. Mar 4, 2013 #2


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    Hello Anon481516! :smile:
    Whatever you do to one side of the =, you must do to the other side.

    So when you multiplied that fraction by [COF2]2,

    the [COF2]2 on the bottom of the fraction disappears, but you must multiply the other side of the = by [COF2]2 also …

    since it's multiplication, it appears on the top.​

    By contrast, you are dividing both side by [CF4] …

    so the [CF4] is divided out (and disappears) from the top of the fraction, and the other side of the = must also be divdied by it, so it ends up on the bottom.
    The writer just swapped sides for no particular reason.

    He started with 1 = BC/D, he multiplied both sides by D/C to get D/C = B,

    and then for no particular reason swapped side to get B = D/C. :wink:
    Yes, you just multiply both sides of the equation by one factor, so it disappears from the bottom and reappears on the top on the other side of the =
    Not following you. :confused:
  4. Mar 4, 2013 #3
    Don't worry about that, I was just using it as an example to elaborate on my problem.

    It does make sense the way you did it. However, the way I was trying to do it, was trying to isolate the CO2 right at the beginning, trying to move it to the other side. I had similar problems when trying to simplify differential calculus questions. My algebra was always the problem.

    Do you know of any examples with solutions, perhaps detailed explanations? What you said helped, but I'm still not sure why you did it that way. Perhaps its because you have to and those are the rules of math and I can understand that, I just don't understand the rules.

    I feel fine with things like:

    E = I*R and P = I2*R

    But other ones I just forget how to do:

    E = [itex]\sqrt{}P*R[/itex] and I = [itex]\sqrt{}P/R[/itex] <---- Its supposed to be a square root symbol.

    Could you or someone else explain those rules? I know this seems a bit excessive, there is probably many of them, however I'd like to rid myself of this problem after so many years. Is there a small book, an article/post somewhere that describes or explains this?
  5. Mar 4, 2013 #4


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    Did you ever study algebra? That's where most of the rules come from. The rest come from arithmetic.
  6. Mar 5, 2013 #5


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    P = I2*R​

    divide both sides by R …

    P/R = I2

    take the square root of both sides

    √(P/R) = I​

    finally substitute that formula for I in the original E = I*R …

    E = I*R = (√(P/R))*R = (√P)*R/√R = (√P)(√R) = √(P*R) :smile:
  7. Mar 5, 2013 #6
    I didnt follow that third part:


    You went from

    (√(P/R))*R = (√P)*R/√R <--- How did you get here?

    I tried isolating for r^2 in the force of gravity equation:

    F = G*m1*m2/r^2

    r^2 = G*m1*m2

    r = √G*m1*m2

    As well as for m1:

    r^2 = G*m1*m2

    r^2/G*m2 = m1

    Are those correct? Could you or someone else explain the methodology/rules you use? Answering and transposing the questions is great, but I could really use an explanation so that the harder transpositions are more clear.
  8. Mar 5, 2013 #7


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    √(P/R) = (√P)/(√R)

    so ##\sqrt{\frac{P}{R}}*R = \frac{\sqrt{P}}{\sqrt{R}}*R = \frac{\sqrt{P}*R}{\sqrt{R}}##

    i can't tell what you mean if you don't use brackets (after the √) to show how much of what follows is included in the √

    can you write it out again, with brackets?

    (and what happened to the "F" in the first line? :confused:)
  9. Mar 5, 2013 #8
    I got it now, thanks.

    As for my transposition, that's a good question, what did happen to the F...

    In the first transposition, everything was square-rooted, the G*m1*m2 and then that entire term is divided by the F. So:

    r = √(G*m1*m2) / F

    The second one:

    F = G*m1*m2 / r^2

    F*r^2 = G*m1*m2 <--- I brought the r^2 from the bottom to the left.

    F*r^2 / G*m2 = m1 <--- I brought both G and m2 simultaneously to the left.

    Apologies, I somehow omitted the F all-together.

    Are those correct? And could you explain what rules are needed to properly transpose questions that are of the same difficulty or of higher difficulty? I don't know of any hard formulas that I could use as an example to transpose. Could you try one that you know of and explain which rules you use to find a variable of interest?
  10. Mar 5, 2013 #9


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    yes, that's fine :smile:

    there's only two rules …

    for an equation:

    whatever you do to one side, you must do to the other side

    eg multipy both by the same

    add both to the same

    take the reciprocal of both

    take the square-root of both

    take the log of both​

    for a fraction:

    whatever you do to the top you must do to the bottom

    (multiplying and dividing only!)​
  11. Mar 5, 2013 #10
    Thats it? I thought there would have been more to it. Nevertheless, I'll still try and find some harder formulas to transpose for extra practice.

    Thanks for the help.
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