I began to solve this problem by realising that 2ab=2, as ab=1.
I could then set the first equation to the second equation, and divide both sides by b.
2ab=2, bc=2
2ab=bc, 2a=c
I then did the same for the third and fourth equations:
cd=3, 4cd=12
de=4, 3de=12
4cd=3de, 4c=3e
Then I substituted c=2a into 4c=3e:
4(2a)=3e, 8a=3e
e=8a/3
Substituting e=8a/3 into ea=6:
(8a/3)*a=6
(8a^2)/3=6
8a^2=18
a^2=18/8=9/4
a=3/2 or a=-3/2
I could then work down the chain of equations to find each value:
ab=1
3b/2=1 or -3b/2=1
b=2/3 or b=-2/3
bc=2
2c/3=2 or -2c/3=2
c=3 or c=-3
cd=3
3d=3 or -3d=3
d=1 or d=-1
de=4
1e=4 or -1e=4
e=4 or e=-4
There are two sets for the values of the {a, b, c, d, e} real numbers, as two negatives in the second set only multiply to generate positive values:
1. {3/2, 2/3, 3, 1, 4}
2. {-3/2, -2/3, -3, -1, -4}