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Get solution for b, “x + b < constraint” for domain_start < x < domain_end?

Posted 1 year ago
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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

2 Replies

If you cross-post to StackExchange, please link the threads together so people won't spend time working on your problem if it is already solved ...

https://mathematica.stackexchange.com/questions/155091/x-b-constraint-for-domain-start-x-domain-end-solution-for-b-only

Solution found by user Mathe172:

Reduce[ForAll[x, 3 < x < 6, x + b < 50], b]

Gives the following result: (* b <= 44 *)

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