       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