Re: solve question

*To*: mathgroup at smc.vnet.net*Subject*: [mg90995] Re: solve question*From*: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>*Date*: Sat, 2 Aug 2008 03:27:02 -0400 (EDT)*Organization*: The Open University, Milton Keynes, UK*References*: <g6uchj$pdf$1@smc.vnet.net>

Francisco Gutierrez wrote: > Frieds: I have a question, as always I imagine it is quite simple. > > I have made several functions. I give the following example: > > newsubstraction[a_List, b_List]:=Module[{k=((b[[2]]+a[[2]])-(b[[1]]+a[[= > 1]]))/2},{middle[a]-middle[b]-k, middle[a]-middle[b]+k}] > > This function takes two lists of the form {a1,b1} and {a2,b2} and returns a list of identical form. Now, suppose I have an equation with an unknown (of the form {u1,u2}). My question is: how to solve it? > I would like something of the sort: > IMAGINEDSOLVE[newsubstraction[firstlist, UNKNOWN]==secondlist, UNKNOWN] > I would like this IMAGINEDSOLVE to work for newsubstraction and other functions of the same type. Thanks for any help. Something along the lines, mySolve[l1_, l2_, uk_] := Solve[Thread[newsub[l1, uk] == newsub[l2, uk]], uk] should do it. Regards, -- Jean-Marc