Will, Riordan and Howard had a total of 128 buttons. The ratio of Riordan's buttons to Howard's buttons was 4 : 9 at first. Will and Riordan each gave away ¹₂ of their buttons. Given that the three boys had 100 buttons left, how many buttons did Will have in the end?

	Will	Riordan	Howard	Total
Comparing Riordan and Howard at first		4 u	9 u
Before	2 p	2x2 = 4 u	9 u	128
Change	- 1 p	-1x2 = - 2 u		- 28
After	1 p	1x2 = 2 u	9 u	100

Total number of buttons that Will and Riordan gave away
= 128 - 100
= 28

The number of buttons that Riordan had at first is repeated.  Make the number of buttons that Riordan had at first the same.  LCM of 4 and 2 is 4.

1 p + 2 u = 128 - 100
1 p + 2 u = 28
1 p = 28 - 2 u --- (1)

1 p + 2 u + 9 u = 100
1 p + 11 u = 100
1 p = 100 - 11 u --- (2)

(1) = (2)
28 - 2 u = 100 - 11 u
11 u - 2 u = 100 - 28
11 u - 2 u = 72
9 u = 72

1 u = 72 ÷ 9 = 8

Substitute 1 u = 8 into (1).
1 p = 28 - 2 u
1 p = 28 - 2 x 8
1 p = 28 - 16

1 p = 12

Number of buttons that Will had in the end
= 1 p
= 12

Answer(s): 12