Hello, I am trying to solve for example the following equation
a + (b - y) (1 - y) x=0
where x is a variable and y is an unknown parameter.
I use:
SolveAlways[a + (b - y) (1 - y) x == 0, x]
and the solution is:
{{a -> 0, b -> y}, {a -> 0, y -> 1}}
I want only the first solution since y is an unknown parameter and in general doesn't equal 1.
Is there a way to tell the program to treat y as a parameter and not to try and solve it for a specific y?
