Sheep.. Sheep...

Let (x,y,z) = # of (cow, pig, sheep)

Then,

x+y+z = 14

($1000)x + ($200)y + ($100)z = $3900 ---> 10x + 2y + z = 39

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9x+y=25

max x = 2 because if x = 3 , then, y will be a negative but there is a negative number of thing so, x = 2.

Then,

9x + y = 25 --->y=25-18=7

And finally,

z = # of sheep = 14-(x+y) = z ---> z = 14-9 = 5

Final Answer: 5.

cow (c) costs €1000, a pig (p) costs €200 and a sheep (s) costs €100.

c + p + s = 14 ------- eq 1 -------- × 10

10c + 10p + 10s = 140 ------- eq 3

1000c + 200p + 100s = 3900 or 10c + 2p + s = 39 ------- eq 2

10c + 10p + 10s = 140 ------- eq 3 and 10c + 2p + s = 39 ------- eq 2

subtract 8p + 9s = 101 or 9s = 101 – 8p = 99 + (2 – 8p)

Then s = (99/9) + (2 – 8p)/9 assume p = 1, 2, 3, ... to get integer solution

when p = 7 then s = (11) + (2 – 8 × 7)/9 = 11 – 6 = 5 ships, 7 pigs and 2 cows