Advertisement
Help Keep Boards Alive. Support us by going ad free today. See here: https://subscriptions.boards.ie/.
If we do not hit our goal we will be forced to close the site.

Current status: https://keepboardsalive.com/

Annual subs are best for most impact. If you are still undecided on going Ad Free - you can also donate using the Paypal Donate option. All contribution helps. Thank you.
https://www.boards.ie/group/1878-subscribers-forum

Private Group for paid up members of Boards.ie. Join the club.

As simple as A, B, C?

  • 20-11-2004 01:06AM
    #1
    Pigman II
    Closed Accounts Posts: 10,921 ✭✭✭✭


    Here's the equations in my problem.

    A* = A + deltaA

    B* = B + deltaB

    C* = C + deltaC

    A* = B* + B + C* + delta1

    A* = B* + B + C + delta2

    B* = C* + C + delta3

    If it's given that A* AND also any term marked with a 'delta' are constants whose values are known is there any way of solving 'B*' or is it just a case of not enough relevant equations?

    Hope that makes sense. Any help gratefully received..


Comments

  • #2
    Pigman II
    Closed Accounts Posts: 10,921 ✭✭✭✭Pigman II


    Never mind!
    
(i)       A* = A + deltaA

(ii)      B* = B + deltaB
 
(iii)     C* = C + deltaC

(iv)      A* = B* + B + C* + delta1
 
(v)       A* = B* + B + C + delta2

(vi)      B* = C* + C + delta3

Therefore

(iv)+(v)       2A* = 2B* + 2B + (C* + C) + (delta1 + delta2)

From (vi)      B* - delta3 = (C* + C)

Therefore      2A* = 2B* + 2B + (B* - delta3) + (delta1 + delta2)

From (ii)      B = B* - deltaB

Therefore      2A* = 2B* + 2(B* - deltaB) + (B* - delta3) + (delta1 + delta2)

               2A* = 5B* - 2deltaB - delta3 + delta1 + delta2

               2A* + 2deltaB + delta3 - delta1 - delta2 = 5B*

               1/5 (2A* + 2deltaB + delta3 - delta1 - delta2) = B*

               B* = 2A* + 2deltaB + delta3 - delta1 - delta2
                    ---------------------------------------
                                      5


Advertisement