solveset() solutions need extra conditions when other symbols are involved

[user234683]

Suppose we want to solve for 'x', and our equation involves another symbolic parameter 'a', whose value we also don't know. If 'a' has certain values, the equation might not be solvable because it results in 'x' disappearing. Examples:

x**a = 1
x*a = 0
x/a = b

The naive solutions to these equations are only valid if a != 0.
So solveset(x*a, x) should return {0} \ {a} instead of just {0}, similar to what happens for solveset(abs(x) - a, x, S.Reals)

[oscarbenjamin]

I currently get the same:

In [10]: solveset(x*a, x)                                                                                                                                     
Out[10]: {0}

I guess this should be something like:

{0 | a != 0} union something...

Similar issue:

In [16]: solveset(a*x-1, x)                                                                                                                                   
Out[16]: 
⎧1⎫
⎨─⎬
⎩a⎭