Dear Berend,Got it! Thankyou very much indeed for your kind support and elaborating the issue.I'm Extremely grateful!!! Eliza 

> Subject: Re: [R] solving simultaneous Equations in R
> From: bhh@xs4all.nl
> Date: Thu, 12 Dec 2013 09:01:02 +0100
> CC: r-help@r-project.org
> To: eliza_botto@hotmail.com
> 
> 
> On 11-12-2013, at 23:56, eliza botto <eliza_botto@hotmail.com> wrote:
> 
> > Dear Berend,
> > 
> > Thankyou very much indeed for you reply. By taking help from your previous reply @ http://r.789695.n4.nabble.com/Simultaneous-equations-td2524645.html i was able to generate the following loop for the calculation of x=x[1] and y=x[2]. 
> > 
> > 
> > fun <- function(x) {
> > 
> >                 f <- numeric(length(x))                                                                                 # read as:
> > 
> > 
> > f[1] <- 1-0.514-(gamma(1/x[1])*gamma(2/x[1]-1/x[2]))/(gamma(2/x[1])*gamma(1/x[1]-1/x[2]))
> > 
> >                 f[2] <- 0.57-
> > 
> > (gamma(1/x[1]-1/x[2])/gamma(1/x[1])-3*gamma(2/x[1]-1/x[2])/gamma(2/x[1])+2*gamma(3/x[1]-1/x[2])/gamma(3/x[1]))/(gamma(1/x[1]-1/x[2])/gamma(1
> > 
> > /x[1])-gamma(2/x[1]-1/x[2])/gamma(2/x[1]))
> > 
> >                 f
> > 
> > }
> > 
> > 
> > startx <- c(0.1,0.15) # start the answer search here
> > 
> > answers<-as.data.frame(nleqslv(startx,fun))
> > 
> > answers
> > 
> > What i cant understand is the concept involved for setting "startx". my x[1] should always be smaller than x[2] and they both should be less than 1. how can i demonstrate it to startx command line?
> > 
> 
> I don’t quite understand what you you mean.
> Your starting values obey the restrictions you specify (if that is what you meant).
> The solution vector has all elements > 1. But it is a solution.
> 
> Do you mean that you want a solution satisfying the constraints you mention?
> I cannot tell if that is possible.
> 
> Generally speaking a square system of equations is solved or not.
> 
> Sometimes you can vary the starting values to get a different solution that obeys the specified constraints.
> If these constraints  are necessary you are not solving a system of equations but trying to find a  parameter set that satisfies certain criteria. Together with a criterion (sum of squares of function values for example) you could use an optimizing algorithm (optim, nlmin, constrOptim to name a few).
> 
> Berend
> 
> 
> > thanks for your help. I m grateful.
> > 
> >   Eliza
> > 
> > 
> > 
> > 
> > > Subject: Re: [R] solving simultaneous Equations in R
> > > From: bhh@xs4all.nl
> > > Date: Wed, 11 Dec 2013 12:43:02 +0100
> > > CC: r-help@r-project.org
> > > To: eliza_botto@hotmail.com
> > > 
> > > 
> > > On 11-12-2013, at 12:16, eliza botto <eliza_botto@hotmail.com> wrote:
> > > 
> > > > Dear users of R,
> > > > I'm trying to solve the following 2 equations simultaneously in R for "x" and "y". I couldn't get through due to my limited knowledge of R.
> > > > 
> > > > 3=1-[(x-1)!(2x-y-1)!/(2x-1)!(x-y-1)!] 
> > > > 
> > > > 6={[(x-y-1)!/(x-1)!]-[3(2x-y-1)!/(2x-1)!]+[2(3x-y-1)!/(3x-1)!]}/{[(x-y-1)!/(x-1)!]-[(2x-y-1)!/(2x-1)!]}
> > > > 
> > > > obviously, ! is factorial.
> > > > kindly help me out on it or at least suggest something.
> > > > I'll be extremely grateful.
> > > 
> > > There are several packages that solve a system of equations.
> > > ktsolve, nleqslv, BB, which you can find in CRAN Task views: "Numerical Mathematics” and “Optimization”.
> > > 
> > > You will have to write your equations in standard R notation.
> > > I can’t tell if your system is solvable.
> > > 
> > > Berend
> > > 
> > > > Eliza
> > > > 
> > > > 
> > > > 
> > > > [[alternative HTML version deleted]]
> > > > 
> > > > ______________________________________________
> > > > R-help@r-project.org mailing list
> > > > https://stat.ethz.ch/mailman/listinfo/r-help
> > > > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> > > > and provide commented, minimal, self-contained, reproducible code.
> > >
> 
 		 	   		  
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