Answer:
5 cows of 5$, 90 sheep of 45$ and 5 horses of 50$.
Explanation:
Cost of 1 cow = $1
Cost of 1 sheep = $0,5
Cost of 1 horse = $10
Let number of cows be x and , number sheep be y and number of horses be z.
...(1)


..(2)
Since , there should be minimum 1 animal. we will be using hit and trial method for the solution:
1) Put value of x =1 , in (1) and (2),we get two equation in two variables:


Solving above equation we get , y =93.79 , z= 5.21 (Not possible)
2) Put value of x = 2 , in (1) and (2),we get two equation in two variables:


Solving above equation we get , y =92.85 , z= 5.15 (Not possible)
3) Put value of x = 5 , in (1) and (2), we get two equation in two variables:


Solving above equation we get , y =90 , z= 5 (possible)
5 cows of 5$, 90 sheep of 45$ and 5 horses of 50$.