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Thread: Rearranging formulas

  1. #1
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    Rearranging formulas

    a = 5b - 2c(3+b) + c make b the subject

    a = 5b - 6c + b + c

    b = 5b + 6c^2 / a

    can I not do that?
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Rearranging formulas

    I would write:
    $\displaystyle a=5b-2c(3+b)+c$
    $\displaystyle \Leftrightarrow a=5b-6c-2bc+c$
    $\displaystyle \Leftrightarrow a-5b+5c=-2bc$
    $\displaystyle \Leftrightarrow b=...$
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  3. #3
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    Re: Rearranging formulas

    Quote Originally Posted by Siron View Post
    I would write:
    $\displaystyle a=5b-2c(3+b)+c$
    $\displaystyle \Leftrightarrow a=5b-6c-2bc+c$
    $\displaystyle \Leftrightarrow a-5b+5c=-2bc$
    $\displaystyle \Leftrightarrow b=...$
    Interesting, but a - 5b + 5c = - 2bc

    why did - 6c become 5c?

    I understand that change the sides change the sign, but why change 6 to 5?

    I have got it now, - 2bc + c change which change -6c to 5c and -2bc + c became - 2bc
    Last edited by David Green; Sep 5th 2011 at 03:20 PM. Reason: Understanding the change
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  4. #4
    MHF Contributor Siron's Avatar
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    Re: Rearranging formulas

    Because:
    -6c+c=-5c
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  5. #5
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    Re: Rearranging formulas

    Quote Originally Posted by David Green View Post

    why did - 6c become 5c?

    I understand that change the sides change the sign, but why change 6 to 5?
    Like terms were added,

    $\displaystyle a = 5b - 2c(3+b) + c$

    $\displaystyle a = 5b - 6c-2bc + c$

    $\displaystyle a = 5b - 6c+c-2bc $

    $\displaystyle a = 5b - 5c-2bc $
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  6. #6
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    Re: Rearranging formulas

    Quote Originally Posted by pickslides View Post
    Like terms were added,

    $\displaystyle a = 5b - 2c(3+b) + c$

    $\displaystyle a = 5b - 6c-2bc + c$

    $\displaystyle a = 5b - 6c+c-2bc $

    $\displaystyle a = 5b - 5c-2bc $
    I have still got to get b on its own, and there is 2 b's?

    So is the rule now to factor the b's out?
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  7. #7
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    Re: Rearranging formulas

    Quote Originally Posted by David Green View Post
    I have still got to get b on its own, and there is 2 b's?

    So is the rule now to factor the b's out?
    Yep, what do you get?
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  8. #8
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    Re: Rearranging formulas

    Quote Originally Posted by pickslides View Post
    Yep, what do you get?
    Not much yet because I am not sure about factoring mixed terms?

    Will have to do some more research on this my course book has poor example of it?
    Last edited by David Green; Sep 5th 2011 at 03:45 PM. Reason: require more information
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  9. #9
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    Re: Rearranging formulas

    I appreciate the honesty

    $\displaystyle a = 5b - 2c(3+b) + c$

    $\displaystyle a = 5b - 6c-2bc + c$

    $\displaystyle a = 5b - 6c+c-2bc $

    $\displaystyle a = 5b - 5c-2bc $

    $\displaystyle a = 5b -2bc - 5c$

    $\displaystyle a = 5\times b -2c\times b - 5c$

    $\displaystyle a = b(5 -2c) - 5c$

    Spoiler:
    $\displaystyle a+5c = b(5 -2c) $

    $\displaystyle b=\frac{a+5c}{5 -2c} $
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  10. #10
    MHF Contributor Siron's Avatar
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    Re: Rearranging formulas

    Sorry David, I didn't see that the first term had also a facor b so what I wrote doesn't make sense, pickslides solution is the good one.
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  11. #11
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    Re: Rearranging formulas

    Quote Originally Posted by Siron View Post
    Sorry David, I didn't see that the first term had also a facor b so what I wrote doesn't make sense, pickslides solution is the good one.
    Thanks for all your help Pickslides and Siron, in alphabetical order?

    Very much appreciated, I will work on it later after work

    Thanks

    David
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  12. #12
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    Re: Rearranging formulas

    Quote Originally Posted by pickslides View Post
    I appreciate the honesty

    $\displaystyle a = 5b - 2c(3+b) + c$

    $\displaystyle a = 5b - 6c-2bc + c$

    Hi Pickslides

    I have just realised what the spolier was you installed, however I have been looking at this today and came up with this;

    a = 5b - 2c(3+b) + c First I loose the bracket
    a = 5b - 6c - 2bc + c Not sure why but in a previous explanation Siron removed a C, so I have followed the same line of reasoning here.
    a = 5b - 5c - 2b + c Now the whole point of this is to get "b" on it's own, and I still have 5b and another - 2b, so;
    a = 5b - 2b - 5c + c
    a = 3b - 5c + c Now I will subtract "C" from both sides
    a - c = 3b - 5c From this point forwards I can't get "b" on it's own, so I need to go back to line "3" from the top and start again;

    a = 5b - 5c - 2b + c Now this time I will try and subtract 1b
    a = 4b - 5c - b + c Now I can subtract "c" from the RHS
    a - c = 4b - 5c - b Now I can subtract "b" from the RHS
    a - c + b = 4b - 5c Now I can divide both sides by "a - c"

    b = 4b - 5c / a - c

    How have I done working that lot out, since I knew nothing really last night?

    Thanks again for all your help to all who helped and advised

    Thanks

    David

    $\displaystyle a = 5b - 6c+c-2bc $

    $\displaystyle a = 5b - 5c-2bc $

    $\displaystyle a = 5b -2bc - 5c$

    $\displaystyle a = 5\times b -2c\times b - 5c$

    $\displaystyle a = b(5 -2c) - 5c$

    Spoiler:
    $\displaystyle a+5c = b(5 -2c) $

    $\displaystyle b=\frac{a+5c}{5 -2c} $
    Thanks

    David
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  13. #13
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    Re: Rearranging formulas

    You can't just remove the 'c'. This has made your working of the problem incorrect.
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  14. #14
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    Re: Rearranging formulas

    Quote Originally Posted by David Green View Post
    Thanks

    David
    Hi Pickslides

    Please advise what you did to loose a C in your first line?

    The original equation was;

    a = 5b - 2c (3 + b) + C

    In my first line of working out I got;

    a = 5b - 6c - 2bc + c

    in your first line of working out you got;

    a = 5b - 6c -2bc

    what did you do with the latter C ?

    The final answer I am sure you had;

    b = 4b - 5c / a - c

    what changed?

    Thanks

    David
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  15. #15
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    Re: Rearranging formulas

    Post PM

    From this step:

    $\displaystyle a = 5b -2bc - 5c$

    We need to isolate $\displaystyle b$ by factoring, recall $\displaystyle ab+ab = a(b+c)$ so

    $\displaystyle a = 5\times b -2c\times b - 5c$

    $\displaystyle a = b(5 -2c) - 5c$

    $\displaystyle a+5c = b(5 -2c) $

    Now divide $\displaystyle 5-2c$ from both sides

    $\displaystyle b=\frac{a+5c}{5 -2c} $

    All good?
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