Please Help!(not sure what else to title it)

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shadeshigeru

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Feb 14, 2010
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[A+(D*2)]+[B+(D*2)]=X=(C*E)+(D*E)*2
(C*R)+(D*R)*2=[A+(D*2)]
(C*W)+(D*W)*2=[B+(D*2)]
[A+(D*2)]=Y
[B+(D*2)]=Z
Y=593
Z=367
X=960
E=R+W
R and W must be whole numbers
I'm stumped here solving for all unknowns. I believe there can be more than one solution, i'm only looking for one solution, if anybody here could help i'd appreciate it.
 
shadeshigeru said:
[A+(D•2)]+[B+(D•2)]=X=(C•E)+(D•E)2
Click to expand...
What does this mean: (D•E)2 ? 2 times D times E ?

WHY is there 2 equal signs?

[A+(D•2)]+[B+(D•2)]
That is same as A + B + 4D ; agree?
 
What/how you posted is very confusing...

Can you confirm that the 1st 3 equations can be written this way:

a + b + 4d = ce + 2de = 960, where e = r + w

cr + 2dr = a + 2d = 593

cw + 2dw = b + 2d = 367

AND did you notice that 960 - 593 = 367 ?
 
Yes the first 3 equations can be written like that. However i wrote them how I did for a reason.


Just a heads up, this is not homework help or anything like that. This is a real life problem and i wrote the equation myself in order to try and find out how to solve the real life problem. Solve the equation, solve the real life problem. Yes i am aware that 960-593=367. :)

Each variable stands for an actual real life item. For instance, R stands for (number of items in group 1) and W stands for (number of items in group 2) where as D stands for spacing. I hope that helps to clarify why the equation was written how it was.
 
Well, even after restricting the variables a,b,c,d,e to integers, there are still 780 solutions to your problem!
A few (in order a,b,c,d,e,r,w, r+w):
275, 49, 2, 159,3, 1.853125, 1.146875, 3
535, 309, 6, 29, 15, 9.265625, 5.734375, 15
591, 365, 1, 1, 320, 197.67, 122.33, 320

We have these 3 equations:
a + b + 4d = ce + 2de = 960 [1], where e = r + w

cr + 2dr = a + 2d = 593 [2]

cw + 2dw = b + 2d = 367 [3]

[1]: e(c + 2d) = 960
e = 960 / (c + 2d) [4]

[2]: r(c + 2d) = 593
r = 593 / (c + 2d) [5]

[3]: w(c + 2d) = 367
w = 367 / (c + 2d) [6]

[5] + [6] = 593 / (c + 2d) + 367 / (c + 2d) [5] = 960 / (c + 2d) = [4]

In other words, e = r + w is already given; so no help in being specified as a given.

A few "defaults" can be obtained, like:
[2][3]: a + 2d - (b + 2d) = 593-367
leads to: b = a - 226 ; so we know a > 226 (unless zero and negatives are allowed).

[2]: 2d = 593 - a
d = (593 - a) / 2 ; we know a < 592, plus a is odd (division by 2) if d is an integer.

All the 780 solutions can be obtained by a short looper program; condensed:
Loop a from 227 to 591 step 2
Using a, calculate b and d
Loop c (keeping c + 2d =< 367)
Using a,b,d,c, calculate e,r,w
 
Wow thanks for the help. Unfortunately it appears i did not finish the equation when i wrote it so the problem is not solved. Can you re-evaluate it with these additional rules?


D<50
C>90
E<20


Thanks a bunch
 
shadeshigeru said:
Wow thanks for the help. Unfortunately it appears i did not finish the equation when i wrote it so the problem is not solved. Can you re-evaluate it with these additional rules?


D<50
C>90
E<20


Thanks a bunch
Click to expand...

Looks to me like several people put a lot of time and effort into the answers they've given. If you failed to include those "additional rules," I think you should be the one who re-evaluates with these additional rules.
 
Looks to me like several people put a lot of time and effort into the answers they've given. If you failed to include those "additional rules," I think you should be the one who re-evaluates with these additional rules.
Click to expand...


Thanks for doing nothing to contribute to the post... that's helpful. I'm obviously having issues with it, otherwise i would not have posted on this forum... I do appreciate the work that he put in, however i am still at a loss for how to solve it with the additional rules, i don't appreciate your lack of effort yet feeling that you had to contribute your 2 cents. Next time, just ignore the post if you can't contribute yeah?
 
Well, me no wanna get in the middle, so I'll END my involment by telling
you that these 3 restrictions reduce the number of solutions to 197.
 
shadeshigeru said:
Can you give me one of those solutions? Dude is just a troll, i still need help.
Click to expand...
The "Dude" is a Lady, with close to 1500 posts here; call that a troll ? :shock:
 
Well she has obviously succeeded in keeping the purpose of this thread from progressing any further. So absolutely yes. Now i have to go find help somewhere else, such a hassle caused by someone that couldn't just not contribute an opinion that had nothing to do with helping to solve the problem at hand.
 
shadeshigeru said:
Well she has obviously succeeded in keeping the purpose of this thread from progressing any further.

To me, it seems like you are your own roadblock.
Click to expand...

If r and w are restricted to the set of Whole numbers, then why did you not tell Denis that those decimal values are not valid?

Something doesn't jive, here. 8-)
 
Because after he solved it i figured i can deal with them not being whole numbers.

As stated before, this is not a homework problem or anything like that, it's a real-life problem trying to solve in order to help me get a real life solution to the reason i wrote the equation in the first place. I'm not a math person, I use it for my job and that's about the extent of it.
 
shadeshigeru said:
R and W must be whole numbers

I'm assuming that you understand usage of the auxiliary verb "must" imposes an imperative requirement. In other words, either r and w are restricted to the set of Whole numbers or they are not.
Click to expand...

With respect only to the system of equations for which you would like help (i.e., the one that you posted), is the Whole-numbered requirement for r and w no longer true?
 
I'd prefer if they were whole numbers, but as long as E is a whole number i can deal. It would be IDEAL if R or W one were equal to 3
 
I could not find a solution (working by hand) that matches the following constraints.

E must be a Whole number less than 20

D must be less than 50

C must be greater than 90

However, if we change the restriction on E from a Whole number to an Integer (i.e., allow negative values), then we have the following solution.

A = 685

B = 459

C= 91

D = -46

E = -960

R = -593

W = -367

I solved the system:

for A and B in terms of D,

for R and E in terms of W,

and for C in terms of D and W.

D = D

W = W

A = 593 - 2D

B = 367 - 2D

C = 367/W - 2D

E = 960W/367

R = 593W/367

Looking at E: in order for 367 to divide evenly into 960W, W must be a multiple of 367, since 367 is a Prime number. Therefore, the only way that E can be less than 20 is if W is negative.

And, if W is negative, then the only way to get C to be greater than 90 (positive) is to make D negative.

Does the solution above (in blue) work for you? 8-)
 
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