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Homework Help: Subject of the formula

  1. Feb 27, 2014 #1
    I've currently got a quick and short question about making something the subject of the formula. I've managed to do most of it myself, I'm just stuck on the last bit.

    Here is a picture of what I've got so far, how do I make 't' the subject of the formula?

    Here is a picture of the formula: http://puu.sh/7chNU.jpg [Broken]

    Thanks!
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 27, 2014 #2

    adjacent

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    Show your effort first.PF rules requires it.
    btw,This is a homeWork question
     
  4. Feb 27, 2014 #3
    These are my efforts, there were two equations and I made them into one. I just need assistance on how I should take the -3 over onto the other side. It's a simple question that requires a one answer response really.
     
  5. Feb 27, 2014 #4

    DataGG

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    What's this "subject of the formula"? I've never heard of such a thing.. Is it isolating the ##t## in the equation?
     
  6. Feb 27, 2014 #5
    Yes, we are trying to make 't' the subject of the formula by moving everything onto the other side. I've managed to move every other equation onto the other side, but I'm stuck on how to move the -3 over and where it'll end up at.
     
  7. Feb 27, 2014 #6

    adjacent

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    To make things easier,Let's convert the Question to LaTeX,
    ##\sqrt\frac{3}{S+4}=-3t##

    Simply move it.That's all.
    For example,
    y=-3t
    What will you do here?
     
    Last edited: Feb 27, 2014
  8. Feb 27, 2014 #7
    I understand that we have to divide by -3 to move it onto the other side, but I don't exactly know where it'll then end up once we've done so, that's the issue I have.
     
  9. Feb 27, 2014 #8

    adjacent

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    What?
    In my above example,
    ##y=-3t##
    so ##t=\frac{y}{-3}##
    Do the same for your equation
     
  10. Feb 27, 2014 #9
    If I do the same as your example it'll be:

    √(3/S+4-3) = t - Was it that simple?
     
  11. Feb 27, 2014 #10

    adjacent

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    That's wrong.
    Show your steps,to see where you went wrong.
     
  12. Feb 27, 2014 #11
    Basically you stated that y = -3t, so when we divide the -3 from t it'll equal y/-3 = t.
    That's what I attempted to do with the formula above, unless it'll go out from the square root and equal:

    √(3/S+4) / -3

    That's the only other solution I can think of.
     
  13. Feb 27, 2014 #12

    jtbell

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    Neither had I! A Google search shows that it's apparently a British English expression. In American English we usually say something like "solve the equation for t" or "isolate t".
     
  14. Feb 27, 2014 #13

    jtbell

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    Yes, that's it. If you move it inside the square root you have to square it: ##3 = \sqrt{3^2} = \sqrt{9}##.
     
  15. Feb 27, 2014 #14

    adjacent

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    Are these two answers the same?
    BTW,The two answers are is wrong.Do you know why?
    It's because you used brackets in the wrong way.

    Use brackets in the correct way and post the answer.
    Sorry I am not able to give the direct answer because PF rules are against it. :redface:
     
  16. Feb 27, 2014 #15
    I'm going to be honest here, I've stumped and I have no clue what to do.
    The only other way I can think of is:

    √((3/S+4) / -3)
     
  17. Feb 27, 2014 #16

    adjacent

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    Double post...
     
    Last edited: Feb 27, 2014
  18. Feb 27, 2014 #17

    adjacent

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    The original question was this:
    ##\sqrt{\frac{3}{s+4}}=-3t##
    Move -3 to right side.and this gives:
    attachment.php?attachmentid=67088&stc=1&d=1393517679.png
    Simple... :wink:
     

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    Last edited: Feb 27, 2014
  19. Feb 27, 2014 #18
    So this way, √((3/S+4) / -3), was the correct way?
     
  20. Feb 27, 2014 #19

    DataGG

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    DerekBrown, in what grade are you? You should probably ask your teacher for help because basic algebra is absolutely needed in every single math/physics course you'll take in your life.
     
  21. Feb 27, 2014 #20

    DataGG

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    No, that is not the correct way.

    Using adjacent's equation (you'll be remembered adjacent!!!) ##y=-3t##, you know that ##t=y/(-3)##. Now, what happens if ##y## equals the left side of your original equation?
     
  22. Feb 27, 2014 #21

    adjacent

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    *sigh*

    That gives ##\sqrt{\frac{\frac{3}{S}+4}{-3}}##
    You know it's wrong,don't you?

    According to BIDMUS or BODMUS(or whatever)
    You have to divide before adding so 3/s is divided first.But the correct way is,S+4 add first.That's where brackets come into play.Brackets are always executed first.B is brackets in BIDMUS rule

    First write in hand then Use BIDMUS rule and use brackets the correct way.You'll get incorrect results if you are not careful with brackets.

    haha.That equation will become more famous than ##E=mc^2## . :devil:
     
    Last edited: Feb 27, 2014
  23. Feb 27, 2014 #22
    If I'm going to be insulted for being confused on these forums then forget it, I'll take my questions else where mmf. Thank you for your assistance mmf.
     
  24. Feb 28, 2014 #23

    adjacent

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    Just follow BIDMUS rule.
    Anyway,since you are only confused about the brackets,this is the correct answer:
    ##\frac{\sqrt{\frac{3}{s+4}}}{-3}=t##
    Now check the differences between your answer(with bracket problem) with this.
    This can be written as:
    (√(3/(s+4)))/(-3)=t

    See how the brackets are used.
     
  25. Feb 28, 2014 #24

    DataGG

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    I didn't insult you. If you think that telling you to ask a professor for assistance is insulting you, then I believe you're really short tempered.

    I didn't mean to say it in a condescending way.. I really think that you should solidify your algebra, and a good way to do that is by having someone with you, face to face, to teach you. Algebra is absolutely necessary.

    But hey, whatever works for you.
     
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