Luis, Jack and Jeremy had a total of 69 coins. The ratio of Jack's coins to Jeremy's coins was 6 : 5 at first. Luis and Jack each gave away ¹₂ of their coins. Given that the three boys had 47 coins left, how many coins did Luis have at first?

	Luis	Jack	Jeremy	Total
Comparing Jack and Jeremy at first		6 u	5 u
Before	2 p	2x3 = 6 u	5 u	69
Change	- 1 p	-1x3 = - 3 u		- 22
After	1 p	1x3 = 3 u	5 u	47

Total number of coins that Luis and Jack gave away
= 69 - 47
= 22

The number of coins that Jack had at first is repeated.  Make the number of coins that Jack had at first the same.  LCM of 6 and 2 is 6.

1 p + 3 u = 69 - 47
1 p + 3 u = 22
1 p = 22 - 3 u --- (1)

1 p + 3 u + 5 u = 47
1 p + 8 u = 47
1 p = 47 - 8 u --- (2)

(1) = (2)
22 - 3 u = 47 - 8 u
8 u - 3 u = 47 - 22
8 u - 3 u = 25
5 u = 25

1 u = 25 ÷ 5 = 5

Substitute 1 u = 5 into (1).
1 p = 22 - 3 u
1 p = 22 - 3 x 5
1 p = 22 - 15

1 p = 7

Number of coins that Luis had at first
= 2 p
= 2 x 7
= 14

Answer(s): 14