Caden, Justin and Xavier had a total of 46 coins. The ratio of Justin's coins to Xavier's coins was 4 : 3 at first. Caden and Justin each gave away ¹₂ of their coins. Given that the three boys had 29 coins left, how many coins did Caden have at first?

	Caden	Justin	Xavier	Total
Comparing Justin and Xavier at first		4 u	3 u
Before	2 p	2x2 = 4 u	3 u	46
Change	- 1 p	-1x2 = - 2 u		- 17
After	1 p	1x2 = 2 u	3 u	29

Total number of coins that Caden and Justin gave away
= 46 - 29
= 17

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 4 and 2 is 4.

1 p + 2 u = 46 - 29
1 p + 2 u = 17
1 p = 17 - 2 u --- (1)

1 p + 2 u + 3 u = 29
1 p + 5 u = 29
1 p = 29 - 5 u --- (2)

(1) = (2)
17 - 2 u = 29 - 5 u
5 u - 2 u = 29 - 17
5 u - 2 u = 12
3 u = 12

1 u = 12 ÷ 3 = 4

Substitute 1 u = 4 into (1).
1 p = 17 - 2 u
1 p = 17 - 2 x 4
1 p = 17 - 8

1 p = 9

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

Answer(s): 18