Dylan, Justin and Sam had a total of 33 coins. The ratio of Justin's coins to Sam's coins was 2 : 3 at first. Dylan and Justin each gave away ¹₂ of their coins. Given that the three boys had 21 coins left, how many coins did Dylan have at first?

	Dylan	Justin	Sam	Total
Comparing Justin and Sam at first		2 u	3 u
Before	2 p	2x1 = 2 u	3 u	33
Change	- 1 p	-1x1 = - 1 u		- 12
After	1 p	1x1 = 1 u	3 u	21

Total number of coins that Dylan and Justin gave away
= 33 - 21
= 12

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

1 p + 1 u = 33 - 21
1 p + 1 u = 12
1 p = 12 - 1 u --- (1)

1 p + 1 u + 3 u = 21
1 p + 4 u = 21
1 p = 21 - 4 u --- (2)

(1) = (2)
12 - 1 u = 21 - 4 u
4 u - 1 u = 21 - 12
4 u - 1 u = 9
3 u = 9

1 u = 9 ÷ 3 = 3

Substitute 1 u = 3 into (1).
1 p = 12 - 1 u
1 p = 12 - 1 x 3
1 p = 12 - 3

1 p = 9

Number of coins that Dylan had at first
= 2 p
= 2 x 9
= 18

Answer(s): 18