RE: [Q] Equation solving?
- To: mathgroup at smc.vnet.net
- Subject: [mg23318] RE: [mg23282] [Q] Equation solving?
- From: "David Park" <djmp at earthlink.net>
- Date: Mon, 1 May 2000 18:00:18 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
> From: James [mailto:research at proton.csl.uiuc.edu] To: mathgroup at smc.vnet.net > Hello! > > I'm a newbie to Mathematica > and have a question about what's wrong in the following > expression. > (The parameters are generated by the program, > so there're some terms that can be cancelled out.) > > Solve[{ Q1==(2*z+1-1)^2+4*1*1 , > k1==(1/Sqrt[Q1])*(v*z-(3/2)*(2*z+1+1-Sqrt[Q1])) , > Q2==(2*z+1-1)^2+4*1*1 , > k2==(1/Sqrt[Q2])*((6.10-v)* z-(2/2)*(2*z+1+1-Sqrt[Q2])) , > (3/2)*(2*z+1+1)+(k1-(3/2))*Sqrt[Q1] == > 6.10*z-((2/2)*(2*z+1+1)+(k2-(2/2))*Sqrt[Q2])}, {z,v}] > > I want to solve the above equations according to 'z', > and I tried several ways in expressing {z,v} or z, {v}, etc > to get an answer(s), but all in vain. > > Is there anyone who can tell me how to express the equations? > Sorry about the question, if this is trivial. > Thank you. > > > James. > This seems to work: eqns = {Q1 == (2*z + 1 - 1)^2 + 4*1*1, k1 == 1/Sqrt[Q1]*(v*z - 3/2*(2*z + 1 + 1 - Sqrt[Q1])), Q2 == (2*z + 1 - 1)^2 + 4*1*1, k2 == 1/Sqrt[Q2]*((6.1 - v)*z - 2/2*(2*z + 1 + 1 - Sqrt[Q2])), 3/2*(2*z + 1 + 1) + (k1 - 3/2)*Sqrt[Q1] == 6.1*z - (2/2*(2*z + 1 + 1) + (k2 - 2/2)*Sqrt[Q2])}; Eliminate[eqns, {Q1, Q2}] Solve[%, {k1, k2, z}] -3.*k1 + k1^2 == 0. && -2.*k2 + k2^2 == 0. && z == 0. {{z -> 0., k1 -> 0., k2 -> 0.}, {z -> 0., k1 -> 0., k2 -> 2.}, {z -> 0., k1 -> 3., k2 -> 0.}, {z -> 0., k1 -> 3., k2 -> 2.}} Or... Eliminate[eqns, {k1, k2}] Solve[%, {Q1, Q2, z}] Q1 == 4. && Q2 == 4. && z == 0. && Q1 != 0. && Q2 != 0. {{Q1 -> 4., Q2 -> 4., z -> 0.}} It appears that v drops out of the equations. David Park djmp at earthlink.net http://home.earthlink.net/~djmp/