Tommy, Caden and Brandon had a total of 26 coins. The ratio of Caden's coins to Brandon's coins was 4 : 5 at first. Tommy and Caden each gave away ¹₂ of their coins. Given that the three boys had 18 coins left, how many coins did Tommy have in the end?

	Tommy	Caden	Brandon	Total
Comparing Caden and Brandon at first		4 u	5 u
Before	2 p	2x2 = 4 u	5 u	26
Change	- 1 p	-1x2 = - 2 u		- 8
After	1 p	1x2 = 2 u	5 u	18

Total number of coins that Tommy and Caden gave away
= 26 - 18
= 8

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

1 p + 2 u = 26 - 18
1 p + 2 u = 8
1 p = 8 - 2 u --- (1)

1 p + 2 u + 5 u = 18
1 p + 7 u = 18
1 p = 18 - 7 u --- (2)

(1) = (2)
8 - 2 u = 18 - 7 u
7 u - 2 u = 18 - 8
7 u - 2 u = 10
5 u = 10

1 u = 10 ÷ 5 = 2

Substitute 1 u = 2 into (1).
1 p = 8 - 2 u
1 p = 8 - 2 x 2
1 p = 8 - 4

1 p = 4

Number of coins that Tommy had in the end
= 1 p
= 4

Answer(s): 4