David Gwyn J. answered 11/13/20
Highly Experienced Tutor (Oxbridge graduate and former tech CEO)
two simultaneous equations in two unknowns A and B, the length of two classes in hours
(1) 3A + 5B = 12
(2) 12A + 2B = 21
4 x (1) - (2) to eliminate A
12A + 20B = 48
-
12A + 2B = 21
=
0A + 18B = 27
=> B = 27/18 = 1.5
substitute value of B in (1) to get 3(1.5) + 5B = 12
=> 5B = 12 - 4.5 = 7.5
=> B = 1.5
Hence Class A = Class B = 90 minutes (or 1.5 hours)
Double-check? 3(1.5) + 5(1.5) = 4.5 + 7.5 = 12 and 12(1.5) + 2(1.5) = 18 + 3 = 21 both as required.