Re: Replacement question

*To*: mathgroup at smc.vnet.net*Subject*: [mg74219] Re: [mg74156] Replacement question*From*: Bob Hanlon <hanlonr at cox.net>*Date*: Wed, 14 Mar 2007 03:58:54 -0500 (EST)*Reply-to*: hanlonr at cox.net

Convert the rule to one with an atomic expression on the LHS., i.e., "solve" the replacement rule for one of the other variables, preferably a variable which only occurs in the form to be replaced. So {y+z -> x} goes to either {y -> x-z} or {z -> x-y} (y+z)^2+3(y+z)-y+z /. z\[Rule]x-y x^2 + 4*x - 2*y Bob Hanlon ---- Brian Beckage <Brian.Beckage at uvm.edu> wrote: > I apologize for this very basic question. I understand how to use /. > {} to replace a variable with a more complex expression, e.g., x-> > y+z. Can one move in the opposite direction to replace all > occurrences of y+z with x? myExpression/.{y+z->x} does not seem to > work. > > Thank you for your help, > Brian > > > > -- > ********************************************************************* > Brian Beckage > Department of Plant Biology > University of Vermont > Marsh Life Science Building > Burlington, VT 05405 > ********************************************************************* >