ab = 1
bc = 2
cd = 3
de = 4
ea = 6
To solve this problem, we will solve for each variable individually.
Firstly, we solve for a, finding that a = 1/b.
Then, we solve for b, finding that b = 2/c.
This pattern repeats till we solve for e, getting e = 6/a. We have looped straight back to a, and no longer need to continue in this recursive business.
Putting this all together with substitution, we receive a fraction division problem:
1/(2/(3/(4/(6/a)))) = a
We can begin to simplify from here, using our rules about dividing fractions.
1/(2/(3/(4a/6))) = a
1/(2/(18/4a)) = a
1/(8a/18) = a
18/8a = a
Multiplying both sides by 8a cancels out the division on the left side:
18 = 8a^2
Divide by 8 to isolate a^2:
2.25 = a^2
Square root both sides:
1.5 or 3/2 = a
Great! Now we've solved for one variable, we can solve for numerous others with less algebra and more direct computation. It's like the skeleton key to this problem.
Solving for b:
ab = 1
Divide by a
b = 1/a
Substitute in a
b = 1/(3/2)
Divide the fractions
b = 2/3
Solving for c in a similar manner:
bc = 2
Divide by b
c = 2/b
Substitute in b
c = 2/(2/3)
Divide the fractions
c = 3
Solving for d:
cd = 3
Divide by c
d = 3/c
Substitute in c
d = 3/3
d = 1
Solving for e:
de = 4
Divide by d
e = 4/d
Substitute in d
e = 4/1
e = 4
And there we have it!
a = 3/2
b = 2/3
c = 3
d = 1
e = 4