number of chicken --- x
number of cows ---- y
number of horses ---- 100-x-y
.01x + .10y + .50(100-x-y) = 5
times 100
x + 10y + 50(100-x-y) = 500
x+10y+5000-50x-50y = 500
-49x-40y = -4500
49x + 40y = 4500
We need only integer solutions here,
after a few trial and errors I got
x = 60
y = 39
so 60 chickens --- .60
39 cows and -----3.90
1 horse ---------0.50 for a total of $5.00
Is this from an 1850 textbook?
I need SUPER help! A famer has to buy 100 animals, but can only spend exactly $5.00. He has to buy at least on of each animal a chcken for 0.01$, a cow for 0.10$ and a horse for 0.50$!!!! PLease HElp IT means so much I need it by Midnight tonight or else i don't pass my clas :(
2 answers
yes my teacher assigned it to my seventh grade class, thank you so much for your time!! =)