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Algebra Transposition Problem

  1. Dec 17, 2012 #1
    1. The problem statement, all variables and given/known data

    Transpose:

    [itex] {D \over d} = \sqrt{\frac{f + p}{f - p}}[/itex]

    To make f the subject...


    3. The attempt at a solution

    This is neither homework or coursework, it was a little challenge set by my math tutor.

    To be honest I am lost from the start, can someone just let me know if I am on the right track...

    [itex] ({D \over d})^2 = {f + p \over f - p}[/itex]
     
    Last edited: Dec 17, 2012
  2. jcsd
  3. Dec 17, 2012 #2
    Seems right, manipulate it further and you will get "f" as subject.
     
  4. Dec 17, 2012 #3
    Here is an interesting trick you can use. If each side of an equation is a fraction, the numerator plus the denominator divided by the numerator minus the denominator of one side of the equation is equal to the numerator plus the denominator divided by the numerator minus the denominator of the other side of the equation.
     
  5. Dec 20, 2012 #4
    OK well I am am getting no where to be honest. Could I use the above rule to help me in this problem?

    I dont know if the below is going to help or if i were more on the right tracks before.

    [itex]
    {D \over d} = \sqrt{\frac{f + p}{f - p}}\\
    D(f-p)^2= d\sqrt{f+p}\\

    [/itex]
     
    Last edited: Dec 20, 2012
  6. Dec 20, 2012 #5

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    No.

    That doesn't work at all.

    That's a good first step.

    Now, multiply both sides by d2(f-p). You get the same result by "cross multiplying".

    Now distribute the D2 and d2.

    Continue on ...
     
  7. Dec 21, 2012 #6
    ^^OK thanks for you help its much appreciated...



    So after multiplying both sides as you said, then cancelling out the bits that can be cancelled, am I left with the following...

    [itex]
    (f-p)D^2 = (f+p)d^2
    [/itex]

    If so, the only thing, apart from isolating one of those expressions is to do the following (if its correct)

    [itex]
    fD^2 - pD^2 = fd^2 + pd^2
    [/itex]

    then could I do...

    [itex]
    fD^2 = pD^2 + fd^2 + pd^2 \\
    fD^2 - fd^2 = pD^2 + pd^2 \\
    fD^2 = {d^2 + pD^2 + pd^2 \over f} \\
    (fD^2)f = d^2 + pD^2 + pd^2
    [/itex]

    ?

    Also a question I have; as the problem asks to make 'f' the subject, would f2=... be ok or does it have to be f = .... ?
     
    Last edited: Dec 21, 2012
  8. Dec 21, 2012 #7
    I have kept going on the basis the above is right (probably isn't) but anyway...

    [itex]
    (fD^2)f - pD^2 = d^2 + pd^2 \\
    (fD^2)f - D^2 = {d^2 + pd^2 \over p } \\
    (fD^2)f - D^2 = d^2 + d^2 \\
    (fD^2)f - D^2 = d^4 \\
    f^2D^2 - D^2 = d^4 \\
    f^2 = d^4
    [/itex]
     
  9. Dec 21, 2012 #8
    EDIT:

    OK thanks very much.

    so thats the easy way then?

    Does the way I done it make sense though?
     
    Last edited: Dec 21, 2012
  10. Dec 21, 2012 #9
    Yeah!!
    But as you said what I have done is the easiest method.
     
  11. Dec 21, 2012 #10
    the last line is it "- p" or should it just be "p" next to the fraction as if it were times by?
     
  12. Dec 21, 2012 #11
    Yes! It is that fraction times p.
    I have edited it.
     
  13. Dec 21, 2012 #12

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You are good up to here.

    I can't imagine how you think this is the same! You have added the [itex]d^2[/itex] and divided by f. You can't do that with "[itex]fd^2[/itex]". Since you want to solve for f and already have every instance of "f" on the right side, just factor out f: [itex]f(D^2- d^2)= [/itex]

     
  14. Jan 22, 2013 #13
    I can see now why I can do that, I dont know why I thought I could either!

    Sorry for bumping this back up but can you check my new solution pleas? I have completely started from fresh and used the componendo et dividendo rule to make it a lot quicker and easier.


    [itex]
    {D \over d } = \sqrt{f+p \over f-p} \\
    {D^2 \over d^2} = {f+p \over f-p} \\
    {D^2 + d^2 \over D^2 - d^2} = {f \over p } \\
    f = ({\frac {D^2 + d^2}{D^2 + d^2}}) p \\
    [/itex]


    There were go, could anyone check this for me pleas?
     
    Last edited: Jan 22, 2013
  15. Jan 22, 2013 #14

    Mark44

    Staff: Mentor

    Above is correct; you have a mistake below.
     
  16. Jan 22, 2013 #15
    Is it the way it is bracketed and times by p or is it that now it has moved over it should be the sqaure root of it all times by p?

    EDIT: Just remembered this solution had already been given by Vieeneth T
     
  17. Jan 22, 2013 #16

    Mark44

    Staff: Mentor

    No, that has nothing to do with it. Start with where you were correct, and work through the steps for solving for f. It's pretty simple.
    Since giving complete solutions isn't permitted at this forum, his post has been deleted.
     
  18. Jan 22, 2013 #17
    OK I think it was just a typo, I put a + when it should have been a -, if that is the error that was just a typo :)

    [itex]
    f = ({\frac {D^2 + d^2}{D^2 - d^2}}) p \\
    [/itex]
     
  19. Jan 22, 2013 #18

    Mark44

    Staff: Mentor

    That's better!
     
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