Re: Reduce in Ver 6

*To*: mathgroup at smc.vnet.net*Subject*: [mg85537] Re: [mg85464] Reduce in Ver 6*From*: Carl Woll <carlw at wolfram.com>*Date*: Wed, 13 Feb 2008 04:25:06 -0500 (EST)*References*: <200802111113.GAA11303@smc.vnet.net>

Dana DeLouis wrote: >Hi. I was wondering if anyone knows of a more efficient way to do this. >This issue came up in ver 6.0 with these equations: > >equ = { >a + b + c == 3, >a^2 + b^2 + c^2 < 10, >a^3 + b^3 + c^3 == 15, >a^4 + b^4 + c^4 == 35 >} > >I use Reduce, but 'c is returned as a function of a & b. >What I would like is for c to replace a & b with the appropriate values. > >r = {ToRules[Reduce[equ, {a, b, c}]]} > >{{a -> 1, b -> 1 - Sqrt[2], c -> 3 - a - b}, > {a -> 1, b -> 1 + Sqrt[2], c -> 3 - a - b}, > {a -> 1 - Sqrt[2], b -> 1, c -> 3 - a - b}, > {a -> 1 - Sqrt[2], b -> 1 + Sqrt[2], c -> 3 - a - b}, > {a -> 1 + Sqrt[2], b -> 1, c -> 3 - a - b}, > {a -> 1 + Sqrt[2], b -> 1 - Sqrt[2], c -> 3 - a - b}} > > Use the Reduce option Backsubstitution->True. Carl Woll Wolfram Research >My workaround is rather tough, especially if this was much larger. > >convert = {a -> aa_, b -> bb_, c -> cc_} -> > {a -> aa, b -> bb, c -> 3 - aa - bb}; > >new = r /. convert > >{{a -> 1, b -> 1 - Sqrt[2], c -> 1 + Sqrt[2]}, > {a -> 1, b -> 1 + Sqrt[2], c -> 1 - Sqrt[2]}, > {a -> 1 - Sqrt[2], b -> 1, c -> 1 + Sqrt[2]}, > >Etc.. > >Any help is much appreciated. I hope I'm not overlooking something obvious. > >Dana >ddelouis at gmail.com > > > > >

**References**:**Reduce in Ver 6***From:*"Dana DeLouis" <dana.del@gmail.com>