Lee, Sean and Tommy had a total of 47 erasers. The ratio of Sean's erasers to Tommy's erasers was 2 : 3 at first. Lee and Sean each gave away ¹₂ of their erasers. Given that the three boys had 31 erasers left, how many erasers did Lee have at first?

	Lee	Sean	Tommy	Total
Comparing Sean and Tommy at first		2 u	3 u
Before	2 p	2x1 = 2 u	3 u	47
Change	- 1 p	-1x1 = - 1 u		- 16
After	1 p	1x1 = 1 u	3 u	31

Total number of erasers that Lee and Sean gave away
= 47 - 31
= 16

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

1 p + 1 u = 47 - 31
1 p + 1 u = 16
1 p = 16 - 1 u --- (1)

1 p + 1 u + 3 u = 31
1 p + 4 u = 31
1 p = 31 - 4 u --- (2)

(1) = (2)
16 - 1 u = 31 - 4 u
4 u - 1 u = 31 - 16
4 u - 1 u = 15
3 u = 15

1 u = 15 ÷ 3 = 5

Substitute 1 u = 5 into (1).
1 p = 16 - 1 u
1 p = 16 - 1 x 5
1 p = 16 - 5

1 p = 11

Number of erasers that Lee had at first
= 2 p
= 2 x 11
= 22

Answer(s): 22