Luke, Reggie and Billy had a total of 37 coins. The ratio of Reggie's coins to Billy's coins was 2 : 5 at first. Luke and Reggie each gave away ¹₂ of their coins. Given that the three boys had 26 coins left, how many coins did Luke have at first?

	Luke	Reggie	Billy	Total
Comparing Reggie and Billy at first		2 u	5 u
Before	2 p	2x1 = 2 u	5 u	37
Change	- 1 p	-1x1 = - 1 u		- 11
After	1 p	1x1 = 1 u	5 u	26

Total number of coins that Luke and Reggie gave away
= 37 - 26
= 11

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

1 p + 1 u = 37 - 26
1 p + 1 u = 11
1 p = 11 - 1 u --- (1)

1 p + 1 u + 5 u = 26
1 p + 6 u = 26
1 p = 26 - 6 u --- (2)

(1) = (2)
11 - 1 u = 26 - 6 u
6 u - 1 u = 26 - 11
6 u - 1 u = 15
5 u = 15

1 u = 15 ÷ 5 = 3

Substitute 1 u = 3 into (1).
1 p = 11 - 1 u
1 p = 11 - 1 x 3
1 p = 11 - 3

1 p = 8

Number of coins that Luke had at first
= 2 p
= 2 x 8
= 16

Answer(s): 16