I have $ $100.00 and need to buy 100 animals exactly. Also need to buy at least one each. Cows are $10.00. Pigs are $3.00 and chickens are $.50
number of cows --- x
number of pigs ---- y
number of chicks --- 100-x-y
cost relation:
10x + 3y + (1/2)(100-x-y) = 100
times 2:
20x + 6y + 100 - x - y = 200
19x + 5y = 100
y = (100 - 19x)/5
since x and y must both be whole numbers, 100 - 19x must be divisible by 5 and since 100 is divisible by 5 , 19x must be a multiple of 5
Furthermore, 100 - 19x ≥ 0
19x ≤ 100
so 0 ≤ x ≤ 5
The only value of x in that domain which makes 19x
a multiple of 5 is x = 5
if x = 5, y = (100 - 95)/5 = 1, and 100-x-y = 94
So he bought 5 cows, 1 pig and 94 chicken
check:
5+1+94 = 100, so he has 100 animals
cost = 10(5) + 3(1) + 94(1/2) = 100
So he spent all of the $100
my answer is correct