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 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
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Posted 1 year ago
Posted 1 year ago
 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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