Let a, b, c be reals satisfying 3 a b+2=6 b, 3 b c+2=5 c, 3 c a+2=4 a Let Q denote the set of all rational numbers. Given that the product a b c can take two values r s Q and t u Q , in lowest form, find r+s+t+u .

Step-by-step solution

array ll 3 a b+2=6 b ( i ), & 3 b c+2=5 c ( ii ), 3 a c+2=4 a . 3 a b c+2 c=6 b c & (1) 3 a b c+2 a=5 a c & (2) 3 a b c+2 b=4 a b & (3) array Substitute b c from (ii) in (1), aligned & 3 a b c+2 c=10 c-4 & 3 a b c=8 c-4 ...(4) aligned Substitute ac from (iii) in (2), aligned & 3 a b c+2 a=5 ( 4 a-2 3 ) & 9 a b c+6 a=20 a-10 & 9 a b c=14 a-10 ....(5) aligned Substitute a b from (i) in (3), aligned & 3 a b c+2 b=4 [ 6 b-2 3 ] & 9 b a c+6 b=24 b-8 & 9 a b c=18 b-8 ...(6) aligned From (4), (5), (6), aligned & 24 c-12=1