tomtom
#1
in the following, `c`

**should be assumed non-zero** for the solution `x = 1 /c `

:

```
import SymPy
const sympy = SymPy.sympy
julia> eqn = sympy.sympify("x * c - 1")
c⋅x - 1
julia> sol = sympy.solve(eqn, SymPy.Sym("x"))
1-element Array{SymPy.Sym,1}:
1/c
```

the problem is, how could I **automatically get this assumption** upon calling `sympy.solve()`

?

thanks.

I don’t think even `sympy.py`

does that:

```
>>> solve(c * x - 1, x)
[1/c]
```

tomtom
#3
u mean, `SymPy`

could **not** tell me the assumptions needed?

So it looks. You may want to ask the python sympy community if there is a solution for your problem.

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tomtom
#5
thanks.

then a dumb question: which/where is the most active Python SymPy discussion group?

You can find the pointers for the mailing list and repository here.

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