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 | 1 | initial version |
If you are sure that there is only one free variable, then you can easily identify it.
First, use solution_dict option to deal with the result more easily:
sage: sol1 = solve([x+y==1], x, y, solution_dict=True)
sage: sol1
[{y: r3, x: -r3 + 1}]
Now, you can ask for the variables appearing in a symbolic expression, see:
sage: sol1[0][x]
-r3 + 1
sage: sol1[0][x].variables()
(r3,)
sage: sol1[0][x].variables()[0]
So, you can give it a Python name and use it afterwards:
sage: freevar = sol1[0][x].variables()[0]
sage: expr = sol1[0][x]
sage: expr
-r3 + 1
sage: expr.subs({freevar:10})
-9
| | 2 | No.2 Revision |
If you are sure that there is only one free variable, then you can easily identify it.
First, use solution_dict option to deal with the result more easily:
sage: sol1 = solve([x+y==1], x, y, solution_dict=True)
sage: sol1
[{y: r3, x: -r3 + 1}]
Now, you can ask for the variables appearing in a symbolic expression, see:
sage: sol1[0][x]
-r3 + 1
sage: sol1[0][x].variables()
(r3,)
sage: sol1[0][x].variables()[0]
So, you can give it a Python name and use it afterwards:
sage: freevar = sol1[0][x].variables()[0]
sage: expr = sol1[0][x]
sage: expr
-r3 + 1
sage: expr.subs({freevar:10})
-9
