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 2009-07-18, 08:19 #1 Unregistered   23×13×61 Posts need help here is the problem i'm haveing with N=(C+P)*(D+Q) i'm trying 2 get C and D on 1 side but 4 the life of me i can not remember how 2 do it .
 2009-07-18, 12:40 #2 Mini-Geek Account Deleted     "Tim Sorbera" Aug 2006 San Antonio, TX USA 11×389 Posts Remember the Distributive Property. http://en.wikipedia.org/wiki/Distributivity http://www.algebrahelp.com/lessons/s.../distribution/
2009-07-18, 15:34   #3
Primeinator

"Kyle"
Feb 2005
Somewhere near M52..

11100101012 Posts

Quote:
 Originally Posted by Unregistered here is the problem i'm haveing with N=(C+P)*(D+Q) i'm trying 2 get C and D on 1 side but 4 the life of me i can not remember how 2 do it .
Just the term C*D on one side or all terms with C and D by themselves on one side? The second of these scenarios is more difficult.

 2009-07-19, 02:15 #4 Unregistered   17×29 Posts thank 4 the help here is what i have . N=(C+P)*(D+Q) N=CD+CQ+PD+PQ N-PQ=CD+CQ+PD (N-PQ)/Q=CD+C+PD ((N-PQ)/Q)/P=CD+C+D i hope it is right . i'm sorry this is not 4 homework . this is 4 a factoring program i have made. I think that i have found a new way of finding factors . i have look 4 it but i could not find it anywhere.
2009-07-19, 02:59   #5
axn

Jun 2003

34×67 Posts

Quote:
 Originally Posted by Unregistered i hope it is right .
Not even close.

Last fiddled with by axn on 2009-07-19 at 02:59

2009-07-19, 03:16   #6
Mini-Geek
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

11×389 Posts

Quote:
 Originally Posted by Unregistered N-PQ=CD+CQ+PD (N-PQ)/Q=CD+C+PD ((N-PQ)/Q)/P=CD+C+D i hope it is right .
No, it's not right. When dividing, you must divide all terms on both sides by the same value. You could say:
N-PQ=CD+CQ+PD
(N-PQ)/Q=(CD+CQ+PD)/Q
N/Q-P=CD/Q+C+PD/Q
But it does not easily simplify any further. Having the + between the different multiplications on the right side means that the terms are completely separate from each other and must be treated as such. You can't divide Q from the whole left side and from only one term on the right side just because it would help your attempt at simplification.
Quote:
 Originally Posted by Unregistered i'm sorry this is not 4 homework . this is 4 a factoring program i have made. I think that i have found a new way of finding factors . i have look 4 it but i could not find it anywhere.
Hate to burst your bubble, but I seriously doubt this is a way to find factors. Your inability to do some of the most basic algebra strongly suggests you have found nothing new. Now, maybe you rediscovered something. That'd be great too! Figuring something out on your own can be a lot of fun, even if it's already known.

By the way, the equation can easily be solved for C:
N=(C+P)*(D+Q)
N=CD+CQ+PD+PQ
N-PD-PQ=CD+CQ=C(D+Q)
C=(N-PD-PQ)/(D+Q)
C=(N-P(D+Q))/(D+Q)
or D:
N=(C+P)*(D+Q)
N=CD+CQ+PD+PQ
N-CQ-PQ=D(C+P)
D=(N-CQ-PQ)/(C+P)
D=(N-Q(C+P))/(C+P)
But you can't really group the "D and C" terms on one side, unless I'm misunderstanding what you're trying to do.

Last fiddled with by Mini-Geek on 2009-07-19 at 03:16

2009-07-19, 04:07   #7
Primeinator

"Kyle"
Feb 2005
Somewhere near M52..

7×131 Posts

Quote:
 Originally Posted by Mini-Geek No, it's not right. When dividing, you must divide all terms on both sides by the same value. You could say: N-PQ=CD+CQ+PD (N-PQ)/Q=(CD+CQ+PD)/Q N/Q-P=CD/Q+C+PD/Q But it does not easily simplify any further. Having the + between the different multiplications on the right side means that the terms are completely separate from each other and must be treated as such. You can't divide Q from the whole left side and from only one term on the right side just because it would help your attempt at simplification. Hate to burst your bubble, but I seriously doubt this is a way to find factors. Your inability to do some of the most basic algebra strongly suggests you have found nothing new. Now, maybe you rediscovered something. That'd be great too! Figuring something out on your own can be a lot of fun, even if it's already known. What exactly is your idea? By the way, the equation can easily be solved for C: N=(C+P)*(D+Q) N=CD+CQ+PD+PQ N-PD-PQ=CD+CQ=C(D+Q) C=(N-PD-PQ)/(D+Q) C=(N-P(D+Q))/(D+Q) or D: N=(C+P)*(D+Q) N=CD+CQ+PD+PQ N-CQ-PQ=D(C+P) D=(N-CQ-PQ)/(C+P) D=(N-Q(C+P))/(C+P) But you can't really group the "D and C" terms on one side, unless I'm misunderstanding what you're trying to do.
Obtaining a solution for C and D by themselves on one side is impossible- at least in terms of solving explicitly (term CD). However, provided you make some substitutions, this looks similar to a problem I had in Differential Equations...

It could be possible to let n = (C+P)*(D+Q) become y(nth order) = cd + cq + pd + pq. From here it might be possible to solve implicitly for all terms c and d though surely not easy. However, I do not think this is the approach best suited for you.

 2009-07-19, 14:45 #8 Unregistered   2×461 Posts thank 4 the help . I will see if i can use them in my program. here is the way i found factors righ$(N,L) = right$(P,L) * right\$(Q,L) it works . the right of n = right of pq 4 a fix lenght
2009-07-22, 17:27   #9
R.D. Silverman

"Bob Silverman"
Nov 2003
North of Boston

165108 Posts

Quote:
 Originally Posted by Unregistered here is the problem i'm haveing with N=(C+P)*(D+Q) i'm trying 2 get C and D on 1 side but 4 the life of me i can not remember how 2 do it .

You have more than one problem here.

(1) Inability or unwillingness to write grammatically correct sentences.
If you can't take time time to properly compose your questions, then

(2) A failure to recognize that C and D are ALREADY on one side of the
equation. They are both clearly on the right side. Your question

 2009-07-22, 20:11 #10 Unregistered   248010 Posts [QUOTE=R.D. Silverman;182249]You have more than one problem here. (1) Inability or unwillingness to write grammatically correct sentences. If you can't take time time to properly compose your questions, then a reasonable follow-up question is: Why should anyone help you? "to properly compose your questions" - isn't that a split infinitive? I wish you wouldn't be so abusive to people who ask for help
2009-07-24, 21:35   #11
R.D. Silverman

"Bob Silverman"
Nov 2003
North of Boston

23·937 Posts

[QUOTE=Unregistered;182269]
Quote:
 Originally Posted by R.D. Silverman You have more than one problem here. (1) Inability or unwillingness to write grammatically correct sentences. If you can't take time time to properly compose your questions, then a reasonable follow-up question is: Why should anyone help you? "to properly compose your questions" - isn't that a split infinitive? I wish you wouldn't be so abusive to people who ask for help
Abuse? What abuse? A student comes to a professor for help.
It is quite legitimate for the professor to insist that the questions
be properly posed and to be written in gramatically correct English.
It certainly is legitimate to suggest that if a student wants help that
he/she spend an appropriate amount of time proofreading.

Furthermore, you will find that in GOOD universities, professors often
mark down for bad composition, even when the subject is not directly
language related. This is proper. One should use proper grammar
even when writing a paper for (say) a chemistry or math class. And it
is proper for the professor to deduct for bad composition.

Maybe you should GROW UP and stop looking at an attempt
to enforce academic standards as abuse or a personal attack.

As for splitting infinitives, it is quite proper to do so when it is
done for emphasis.

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