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mathsiscool






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rearranging this equation
posted on: 7/27/2007 1:39:29 AM

Hi i am having difficulty rearranging this equation to get an equation for C. Can someone who knows about this stuff help please.
A = B*(100-D)/100+Bx(C-D)-C
is it possible to get C = ?

Anyone got any ideas how to solve this problem?
cramwit




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rearranging this equation
replied on: 7/27/2007 5:07:12 AM

A = B*(100-D)/100+Bx(C-D)-C

A - B*(100-D)/100 = Bx(C-D)-C

= BxC - BxD - C

A - B*(100-D)/100 + BxD = BxC - C

= (Bx - 1)C

[ A - B*(100-D)/100 + BxD ]/(Bx - 1) = C

i think that is correct
cramwit




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rearranging this equation
replied on: 7/27/2007 5:07:52 AM

A = B*(100-D)/100+Bx(C-D)-C

A - B*(100-D)/100 = Bx(C-D)-C

= BxC - BxD - C

A - B*(100-D)/100 + BxD = BxC - C

= (Bx - 1)C

[ A - B*(100-D)/100 + BxD ]/(Bx - 1) = C

i think that is correct
mathsiscool




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rearranging this equation
replied on: 7/28/2007 1:13:07 AM

thanks cramwit i will try it out.
mathsiscool




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rearranging this equation
replied on: 7/31/2007 12:32:56 AM

i have run into difficulties with this particular formula. I believe that this is one of those formulas where even when only one variable is missing,(B) experimental values of B have to be plugged in to see how they effect the other value and as such no single simple solution for B can be stated. Try rearranging this formula to prove B.
Based on this formula.
[(A/100)*B] + (100-B/100) = C

by plugging in experimental values of B i can determine,its most likely to be around 88.5, given that i already knew that A was 1.4516 and C is 1.40 to 2 decimal places.
I wonder if anyone knows if its possible to rearrange this equation to isolate only B. If not can anyone tell me what it is called when an equation can't be simplified in that way?
cramwit




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rearranging this equation
replied on: 8/3/2007 2:28:20 AM

A = B*(100-D)/100+Bx(C-D)-C
A + C = B * (100-D)/100 + Bx(C-D)
A + C = B [ (100-D)/100 + x(C-D) ]
(A + C)/[(100-D)/100 + x(C-D)] = B

There are equations that are inseparable.

maybe B + sqrt( B + D ) = K

or B + cosine(B) = X
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