# “x + b < constraint” for “domain_start < x < domain_end”. Solution for “b” only according to the set “x” domain

I am giving a really simple example that describes my problem really clear.

Example: "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 one simple output "b<=44" but i still have not been able to get a result without the defined domain variable intruding in the solution (e.g. (b <= 44&& 3 < x < 6)||(44 < b < 47 && 3 < x < 50-b))

FullSimplify[Reduce[ { x + b < 50 , x > 3 , x< 6 }, {b}, Reals, Backsubstitution -> True], x > 3 && x<6]

• CylindricalDecomposition[x + b < 50 && 3 < x < 6, {b, x}] almost gets there. – J. M.'s technical difficulties Sep 5 '17 at 11:52
• Reduce[x + b < 50 && 3 < x < 6, {b,x}] does the same =/ unfortunately like everything i have tried so far they include x in their solutions – Panagiotis Kmd Sep 5 '17 at 11:58
You can use ForAll for this kind of problem:
Reduce[ForAll[x, 3 < x < 6, x + b < 50], b]