School Lounge  

Go Back   School Lounge > Math Forum > Recreational Math
User Name
Password
Register FAQ Members List Calendar Search Today's Posts Mark Forums Read


Reply
 
Thread Tools Search this Thread Display Modes
  #1  
Old 10-06-2006, 03:33 PM
Bill Bill is offline
Junior Member
 
Join Date: Sep 2006
Posts: 4
Default I need help with these equations


Considering the following table ...

a | b | c | d
---------------------
256 | 80 | 4 | 0
704 | 80 | 11 | 7
288 | 512 | 0 | 5
736 | 512 | 7 | 12

Given a pair of values for a and b, I derived the following to solve for
c and d ...

c = ( ( a - 256 ) / 64 ) - ( ( b - 464 ) / 96 )
d = ( ( a - 256 ) / 64 ) + ( ( b - 80 ) / 96 )

But given a pair of values for c and d, I am having difficulty finding a
and b. I came up with the following but it is incorrect. any help would
be appreciated, thank you.

a = ( ( c * 64 ) + 256 ) + ( ( d * 96 ) + 464 )
b = ( ( c * 64 ) + 256 ) - ( ( d * 96 ) + 80 )






Reply With Quote
  #2  
Old 10-06-2006, 03:55 PM
Greg Neill Greg Neill is offline
Junior Member
 
Join Date: Oct 2006
Posts: 1
Default I need help with these equations


"Bill" <Bill@hotmail.com> wrote in message
news:UGuVg.18477$vi3.11646@bignews3.bellsouth.net. ..
Quote:
Considering the following table ... a | b | c | d --------------------- 256 | 80 | 4 | 0 704 | 80 | 11 | 7 288 | 512 | 0 | 5 736 | 512 | 7 | 12 Given a pair of values for a and b, I derived the following to solve for c and d ... c = ( ( a - 256 ) / 64 ) - ( ( b - 464 ) / 96 ) d = ( ( a - 256 ) / 64 ) + ( ( b - 80 ) / 96 ) But given a pair of values for c and d, I am having difficulty finding a and b. I came up with the following but it is incorrect. any help would be appreciated, thank you. a = ( ( c * 64 ) + 256 ) + ( ( d * 96 ) + 464 ) b = ( ( c * 64 ) + 256 ) - ( ( d * 96 ) + 80 )


Try:

a = 32*(c + d) + 128

b = 48*(d - c) + 272


Reply With Quote
  #3  
Old 10-06-2006, 04:23 PM
Bill Bill is offline
Junior Member
 
Join Date: Sep 2006
Posts: 4
Default I need help with these equations

Greg Neill wrote:
Quote:
a = 32*(c + d) + 128 b = 48*(d - c) + 272


Works perfectly! You are brilliant, thank you Greg!
Reply With Quote
  #4  
Old 10-12-2006, 01:40 PM
Bill Bill is offline
Junior Member
 
Join Date: Sep 2006
Posts: 4
Default I need help with these equations

> Try:
Quote:
a = 32*(c + d) + 128 b = 48*(d - c) + 272


Greg,

Could you please explain to me step by step how you arrived at that
answer? Thank you
Reply With Quote
  #5  
Old 10-12-2006, 02:13 PM
Greg Neill Greg Neill is offline
Junior Member
 
Join Date: Oct 2006
Posts: 1
Default I need help with these equations

"Bill" <Bill@hotmail.com> wrote in message
news:8BrXg.41972$tT6.32387@bignews7.bellsouth.net. ..
Quote:
Try: a = 32*(c + d) + 128 b = 48*(d - c) + 272 Greg, Could you please explain to me step by step how you arrived at that answer? Thank you


Sure. I started with your equations:

c = ( ( a - 256 ) / 64 ) - ( ( b - 464 ) / 96 ) (1)
d = ( ( a - 256 ) / 64 ) + ( ( b - 80 ) / 96 ) (2)

(first verifying that they did in fact produce
the correct values for c and d...)

Then I assumed that c and d were now constant
values and solved the pair for a and b:

Start with (1) and rearrange for a:

(a - 256)/64 = c + (b - 464)/96

a - 256 = 64*c + (b - 464)*64/96

a = 64*c + (2/3)*b - 160/3 (3)

Now substitute this expression for a into (2):

d = ([64*c + (2/3)*b - 160/3] - 256)/64 + (b - 80)/96

d = c - 29/6 + b/96 + (b-80)/96

d = c + b/48 - 17/3

Rearrange for b:

b = 48*d - 48*c + 272

b = 48*(d - c) + 272 (4)

Now use this value for b in equation (3):

a = 64*c + (2/3)*[48*(d - c) + 272] - 160/3

a = 32*(c + d) + 128 (5)

And there you have it.




Reply With Quote
Reply


Thread Tools Search this Thread
Search this Thread:

Advanced Search
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

vB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Forum Jump



All times are GMT. The time now is 01:08 PM.


MyLounge.com Site Map
Forum: Cars, Cell Phone, Database, Games, Home Improvement, IT, Music, School, Sports, Web Design, Web Server, Weight Loss

The MyLounge.com forum is intended for informational use only and should not be relied upon and is not a substitute for any advice. The information contained on MyLounge.com are opinions and suggestions of members and is not a representation of the opinions of MyLounge.com. MyLounge.com does not warrant or vouch for the accuracy, completeness or usefulness of any postings or the qualifications of any person responding. Please consult a expert or seek the services of an attorney in your area for more accuracy on your specific situation. Please note that our forums also serve as mirrors to Usenet newsgroups. Many posts you see on our forums are made by newsgroup users who may not be members of MyLounge.com Term of Service

Powered by: vBulletin Version 3.0.7
Copyright ©2000 - 2008, Jelsoft Enterprises Ltd.