Solving a system of equations in W|A

Posted 5 months ago
939 Views
|
8 Replies
|
0 Total Likes
|
 I am a new user and I just want to be pointed in the right direction.I have a system of equationsP=Gt, F=fcG, G=D+F/(1-f)+U, T=P-F, U=(1-f)G, N=D-T+F I want to know how to solve for N in terms oft,f,c,DI tried inputting Solve P=Gt, F=fcG, G=D+F/(1-f)+U, T=P-F, U=(1-f)G, N=D-T+F for N,t,f,c,DBut Wolfram|Alpha said it didn’t understand.Any assistance would be much appreciated.Paul Mason
8 Replies
Sort By:
Posted 5 months ago
 Try this with Wolfram|Alpha  Solve[{P=G*t, F=f*c*G, G=D+F/(1-f)+U, T=P-F, U=(1-f)*G, N=D-T+F},{N,t,f,c,D}] That worked. I posted this. Then I tried to check it again and it failed.Working on finding the problem
Posted 5 months ago
 Thank you Bill. I tried pasting your expression into Wolfram|Alpha. It just said it didn’t understand.Do you know if there is a general reference in Wolfram|Alpha Documentation about how to pose this type of question?
Posted 5 months ago
 Yes. That is what I'm seeing now too.I'm almost certain it worked once and I can't get it to do that again.If I do solve P=G*t,F=f*c*G,G=D+F/(1-f)+U,T=P-F,U=(1-f)*G,N=D-T+F for N it says it does not understandBut if I do solve F=f*c*G,G=D+F/(1-f)+U,T=P-F,U=(1-f)*G,N=D-T+F for N then it works.There are limits on how long a line of input it will accept and it seems that there may be limits on how complicated internal computations may be. I do not know that is what is causing the problem at the moment.Sometimes it is possible to divide a problem into two parts, each of which is small and simple enough to complete, and then combine the results to get your final answerI just tried this Solve[{P=G*t, F=f*c*G, G=D+F/(1-f)+U, T=P-F, U=(1-f)*G, N=D-T+F},N] and it worked! And I closed the tab on the browser, opened again, went to Wolfram|Alpha, pasted that, and it worked again. See if you can reproduce that. I'm still not certain if the form of the result is what you are looking for. I'll look at this a little more and see if I can get it to do better.
Posted 5 months ago
 Thank again Bill.I copied in the version you got to work, to no avail.I will check the syntax of my expressions; also try to break the problem up.Paul
Posted 5 months ago
 I tried a different approach ReplaceAll[N=D-T+F, T->P-F] gives me N=D+2F-P and then ReplaceAll[N=D+2F-P, P->G*t] gives me N=D+2*f-G*t and then ReplaceAll[N=D+2*f-G*t, G->D+F/(1-f)+U] gives me N=-t(D+F/(1-f)+U)+D+2*f What that is trying to do is eliminate one variable at a time and get closer to your desired expression only in terms of t,f,c,DBUT you have things like G=D+F/(1-f)+U F=fcG where G gives you something containing F and F gives you something containing G. Likewise G and U. What I had hoped was I might be able to order the replacements in a way to get a result only in terms of what you wanted.But each of these ReplaceAll seems to work
Posted 5 months ago
 Thanks again, Bill.In a little while, I am going to try manually eliminating the intermediate variables I don’t want, then use Simplify.A reference manual for Wolfram|Alpha would be really useful.Paul
 First let's see if we can make F, G and U independent of each other by using Solve Solve[{F=f*c*G, G=D+F/(1-f)+U, U=(1-f)*G},{F,G,U}] That neatly separates the F, G and U and gives us, among a couple of other things, F=c*D*(f-1)/(c+f-1) G=D*(f-1)/(f*(c+f-1)) U=-D*(f-1)^2/(f*(c+f-1)) We are very fortunate that all those are expressed in terms of the variables you are looking for.Next eliminate the T in your original equation N=D-T+F using your given T=P-F ReplaceAll[N=D-T+F, T->P-F] which gives us N=D+2*F-P and then eliminate the P in that using your given P=G*t ReplaceAll[N=D+2*F-P, P->G*t] which gives us N=D+2*F-G*t Now eliminate the F in that using the result from Solve that we did above ReplaceAll[N=D+2*F-G*t, F->c*D*(f-1)/(c+f-1)] which gives us N=2*c*D*(f-1)/(c+f-1)+D-G*t and then eliminate the G in that using the result from Solve that we did above ReplaceAll[N=2*c*D*(f-1)/(c+f-1)+D-G*t, G->D*(f-1)/(f*(c+f-1))] which finally gives us N=-D*(f-1)*t/(f*(c+f-1))+2*c*D*(f-1)/(c+f-1)+D It seems like there should be a simpler and less error prone way of doing this where Wolfram|Alpha does more of the work in a single step. Perhaps someone else can see a way to do that or perhaps in a day or two you or I might realize a way to do that.Please check all this VERY carefully to make certain I haven't made any mistakes.