Ronan Lamy wrote:
>> -- you just want something to work with. So this cooked-up example
>>
>> h[1] >>> solve([x + a**2 + y/b, 1/x - y + a])
>>
>> indicates you want two of the four variables solved for in terms of
>> the
>>
> How do you know it's two? And how do you know which ones? The only way
> to "resist the temptation to guess" is to use all four.
I'm not sure what you mean "use all four". I'm only accustomed to solving for
two variables from two equations (except for the case of matching
coefficients). But I'm fine with not guessing and having a check that if no
variables are given that there are just as many free variables as there are
equations. But then we can't expect something like
solve(Eq(a*x**2 + b*x + c, 3*x**2 + 2*x - 4), [a,b,c])
>>
> In any case, the dictionary solution cannot work with infinite
> solution sets, so it's not a good choice.
Can't solve just yield an infinite number of dictionaries, each corresponding
to one of the infinite solutions?
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