I am giving a really simple example that describes my problem really clear.
"x + b < 50" for "3 <x < 6
i need a solution set for my coefficient (e.g. for "b") that satisfies all possible values of the defined domain of a variable (e.g. "3 < x < 6") for the given function (e.g. "x + b < 50").
I need to extract the simple output "b<=44" but i still have not been able to get this result without the defined domain variable intruding in the solution:
Reduce[x > 3 && ((x < 6 && b <= 44) || (b > 44 && b < 47 && b + x < 50)), {b,x}]
Produces results with x being parameterized for possible solutions: (b <= 44&& 3 < x < 6)||(44 < b < 47 && 3 < x < 50-b)
The 2 commands below that should normally work without giving a "x" variable in the solution do not produce any results:
Reduce[x + b < 50 && 3 < x < 6, {b}]
or
FullSimplify[Reduce[ { x + b < 50 , x > 3 , x< 6 }, {b}, Reals, Backsubstitution -> True], x > 3 && x<6]
https://mathematica.stackexchange.com/questions/155091/x-b-constraint-for-domain-start-x-domain-end-solution-for-b-only