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Math Help - Rearrange in terms of z

  1. #1
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    Rearrange in terms of z

    Hi,
    I need to rearrange this function in terms of z, I'm struggling a little bit (asterisks are multiplication):

    a*(z-b) = -c*d*(f/(f+z))-(c*g)-h*e*(f/(f+z))-(h*i)

    Any help?
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  2. #2
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    Re: Rearrange in terms of z

    So, is this your equation:
    Code:
               cdf        hef
    a(z-b) = - ... - cg - ... - hi
               f+z        f+z
    Note: multiplication implied (like, hi means h*i)
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  3. #3
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    Re: Rearrange in terms of z

    Yes absolutely, that's more elegant.
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  4. #4
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    Re: Rearrange in terms of z

    Hokay Tom: start by multiplying through by f+z:

    a(z-b)(f+z) = -cdf - cg(f+z) - hef - hi(f+z)

    Do you follow that?
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  5. #5
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    Re: Rearrange in terms of z

    I would start by mulitplying both sides through by f+z . Then you have no nasty fractions. After this multiply out any brackets and then group terms of z and z^2 together, what do you get?
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  6. #6
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    Re: Rearrange in terms of z

    Yep, I follow, how does one group those terms?
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  7. #7
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    Re: Rearrange in terms of z

    Quote Originally Posted by TomByrne View Post
    Yep
    GREAT! Now for the big question: are you SURE the term on left is a(z-b);
    like is that "damn z" really there?
    Asking because we'll be getting "z^2" ... which will spoil my evening !!
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  8. #8
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    Re: Rearrange in terms of z

    Quote Originally Posted by TomByrne View Post
    Yep, I follow, how does one group those terms?
    Expand all the brackets first and Wilmer will show you!
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  9. #9
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    Re: Rearrange in terms of z

    You're absolutely correct, I've screwed something up in the original formula, here is the correct version:

    a*z-(a*b) = -c*d*(f/(f+z))-(c*g)-h*e*(f/(f+z))-(h*i)

    sorry
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  10. #10
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    Re: Rearrange in terms of z

    az = -cdf/(f+z)-cg-hef/(f+z)-hi+ab
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  11. #11
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    Re: Rearrange in terms of z

    Quote Originally Posted by TomByrne View Post
    You're absolutely correct, I've screwed something up in the original formula, here is the correct version:

    a*z-(a*b) = -c*d*(f/(f+z))-(c*g)-h*e*(f/(f+z))-(h*i)

    a*(z-b) = -c*d*(f/(f+z))-(c*g)-h*e*(f/(f+z))-(h*i) : (original)


    sorry
    Huh? That's same thing Tom: a(z-b) = az - ab
    You been drinking?
    AGREE??!!
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  12. #12
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    Re: Rearrange in terms of z

    I try to avoid drinking at work, although considering work is currently a laptop in a van, I suppose a drink wouldn't go astray (although it may not aid my abysmal maths skills).
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  13. #13
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    Re: Rearrange in terms of z

    Well, all the same, I'm no closer to a solution. Can you help me or have I squandered my opportunity?
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  14. #14
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    Re: Rearrange in terms of z

    Quote Originally Posted by TomByrne View Post
    az = -cdf/(f+z) - cg - hef/(f+z) - hi + ab
    Now that you're sober...OK, that's correct.
    Problem/complication is, when multiplying through by f+z, we get az(f+z) = azf + az^2:
    so we have a ye olde quadratic; going through with this mess, simplifying and combining, we'll get:
    ax^2 + x(af - ab + cg + hi) + f(cd + cg + he + hi - ab) = 0 : recognize the quadratic?

    Best way (for me anyway) to handle such a huge mudder is:
    Let u = af - ab + cg + hi
    Let v = f(cd + cg + he + hi - ab)

    Then, using quadratic formula:
    x = [-u +- SQRT(u^2 - 4av)] / (2a)

    Any questions?!

    What you think, Pickslide? You proud of me?
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  15. #15
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    Re: Rearrange in terms of z

    Thanks, great. What are the square brackets in this situation?
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