Some may find my first answer a bit Lispish. More Pythonish :
[[s.lhs()==s.rhs().n() for s in S] for S in solve([eq1, eq2, eq3],[x, y,
z])]
[[x == 0.0675142263037092, y == 0.00748577369629076, z ==
0.00748577369629076],
[x == 0.0833157736962908, y == -0.00831577369629076, z ==
-0.00831577369629076]]
HTH,
--
Emmanuel Charpentier
Le vendredi 22 septembre 2017 13:40:45 UTC+2, Emmanuel Charpentier a écrit :
>
> what's wrong with :
>
> map(lambda S:map(lambda s:s.lhs()==s.rhs().n(), S), solve([eq1, eq2, eq3],
> [x, y, z]))
>
> [[x == 0.0675142263037092, y == 0.00748577369629076, z ==
> 0.00748577369629076],
> [x == 0.0833157736962908, y == -0.00831577369629076, z ==
> -0.00831577369629076]]
>
> Which shows that the first solution fulfills your constraints ?
>
> HTH,
>
> --
> Emmanuel Charpentier
>
> Le jeudi 21 septembre 2017 20:27:43 UTC+2, Natalie Ulrich a écrit :
>>
>> I'm using SageMathCell to solve chemical equilibrium problems, so at
>> least one set of my solutions has to be real and positive.
>>
>> Here's my code:
>>
>> var('x, y, z')
>>
>> xi=0
>>
>> yi=0.150/2.0
>>
>> zi=0.150/2.0
>>
>> K=8.3e-4
>>
>> eq1=K == y*z/ x
>>
>> eq2=xi+yi==x+y
>>
>> eq3=2*xi+2*zi==2*x+2*z
>>
>> solve([eq1, eq2, eq3],[x, y, z])
>>
>>
>>
>> And here are my solutions:
>>
>> [[x == -1/200000*sqrt(2496889) + 15083/200000, y ==
>> 1/200000*sqrt(2496889) - 83/200000, z == 1/200000*sqrt(2496889) -
>> 83/200000], [x == 1/200000*sqrt(2496889) + 15083/200000, y ==
>> -1/200000*sqrt(2496889) - 83/200000, z == -1/200000*sqrt(2496889) - 83/
>> 200000]]
>> ------------------------------
>>
>>
>> Any thoughts? Thanks in advance.
>>
>>
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