Bryan, Howard and Ian had a total of 89 erasers. The ratio of Howard's erasers to Ian's erasers was 8 : 3 at first. Bryan and Howard each gave away ¹₂ of their erasers. Given that the three boys had 55 erasers left, how many erasers did Bryan have at first?

	Bryan	Howard	Ian	Total
Comparing Howard and Ian at first		8 u	3 u
Before	2 p	2x4 = 8 u	3 u	89
Change	- 1 p	-1x4 = - 4 u		- 34
After	1 p	1x4 = 4 u	3 u	55

Total number of erasers that Bryan and Howard gave away
= 89 - 55
= 34

The number of erasers that Howard had at first is repeated.  Make the number of erasers that Howard had at first the same.  LCM of 8 and 2 is 8.

1 p + 4 u = 89 - 55
1 p + 4 u = 34
1 p = 34 - 4 u --- (1)

1 p + 4 u + 3 u = 55
1 p + 7 u = 55
1 p = 55 - 7 u --- (2)

(1) = (2)
34 - 4 u = 55 - 7 u
7 u - 4 u = 55 - 34
7 u - 4 u = 21
3 u = 21

1 u = 21 ÷ 3 = 7

Substitute 1 u = 7 into (1).
1 p = 34 - 4 u
1 p = 34 - 4 x 7
1 p = 34 - 28

1 p = 6

Number of erasers that Bryan had at first
= 2 p
= 2 x 6
= 12

Answer(s): 12