SOLUTION: Gil is twice as old as Bill was, when Gil was 8 years younger than Bill is now. The sum of their ages is 60. C = how old gil is now B = how old Gil was plu 6 A = how old bill w

Gil is twice as old as Bill was, when Gil was 8 years younger than Bill is now. The sum of their ages is 60.
C = how old Gil is now
B = how old Gil was plus 6
A = how old Bill was divided by 4 
What are A, B & C? need for a combination lock puzzle. 
Thanks!

It's a shame you have a Bill and a B too.  I'm going to change his name to
William, so I can use their initials for their ages:

Gil is twice as old as William was, when Gil was 8 years younger than William is now. The sum of their ages is 60.

C = how old Gil is now
B = how old Gil was plus 6
A = how old William was divided by 4 

What are A, B & C? need for a combination lock puzzle.

Gil is twice as old as William was, when Gil was 8 years younger than William is now.

Therefore Gil's age now equals twice William's age minus x years, 

G = 2(W - x)

and 

Gil's age minus x years equals William's age now minus 8 years.

G - x = W - 8

The sum of their ages is 60.

G + W = 60

So we have the system of equations:



Can you solve that system?  If not post again asking how.
The system simplified and in standard form is

G + 2x - 2W =  0
G -  x -  W = -8
G      +  W = 60

solution:  G = 32, x = 12, W = 28

C = how old Gil is now = 32
B = how old Gil was (x=12 years ago) plus 6 = 32-12+6 = 26 
A = how old William was (x=12 years ago) divided by 4 = (28-12)÷4 = 16÷4 = 4

Edwin