solve() only gives one of infinitely many solutions.

Hopefully the screenshot will make my question clear. From what I can tell by looking at the documentation for the solve() function, this application of solve is Supposed to give a parametrization for All solutions to x*y=1, but I only get one.

There are no additional assumptions on my variables x,y. Code in the screenshot was immediately preceded by clearing all variables.

Arthur

Answer 1

[Zoltán Csáti]

Your mistake is that you applied == instead of =. Do this one:

solution = solve('x*y=1','x','y');

[Arthur]

This gives the same answer unfortunately.

Answer 2

[Sean de Wolski]

The problem is that you're asking for a solution for both variables. Instead, solve for one at a time:

solve(x*y==1,x)

ans =
1/y

[Arthur]

Thanks Sean, but what I'm hoping for is a parametrization of all solutions for the variables I list in terms of a new parameter. In the example I give, a solution would look like

x = z

y = 1/z

I'm not interested so much in solving x*y==1, but this was the simplest example I could find that expressed my problem. Furthermore, I assumed that since solve() gives ALL solutions when given a system of polynomials with only finitely many solution (e.g

V = solve(x^2 + x*y + y == 3,x^2 - 4*x + 3 == 0, x, y)

) that there might be some way of parameterizing All solutions in the case where there are infinitely many?

[Sean de Wolski]

That's what it's doing: 1/y is the solution for all x given all y.

If you want to substitute in new values, use subs to do this.

subs(expr,old,new)