Answer:
5 cows of 5$, 90 sheep of 45$ and 5 horses of 50$.
Step-by-step explanation:
Cost of 1 cow = $1
Cost of 1 sheep = $0,5
Cost of 1 horse = $10
Let number of cows be x and , number sheep be y and number of horses be z.
[tex]x+y+z=100[/tex]...(1)
[tex]1\$x+0.5\$y+10\$z=100\$[/tex]
[tex]10x+5y+100z=1000[/tex]
[tex]2x+y+20z=200[/tex]..(2)
Since , there should be minimum 1 animal. we will be using hit and trial method for the solution:
1) Put value of x =1 , in (1) and (2),we get two equation in two variables:
[tex]y+z=99[/tex]
[tex]y+20z=198[/tex]
Solving above equation we get , y =93.79 , z= 5.21 (Not possible)
2) Put value of x = 2 , in (1) and (2),we get two equation in two variables:
[tex]y+z=98[/tex]
[tex]y+20z=196[/tex]
Solving above equation we get , y =92.85 , z= 5.15 (Not possible)
3) Put value of x = 5 , in (1) and (2), we get two equation in two variables:
[tex]y+z=95[/tex]
[tex]y+20z=190[/tex]
Solving above equation we get , y =90 , z= 5 (possible)
5 cows of 5$, 90 sheep of 45$ and 5 horses of 50$.