SOLUTION: A city wants to buy some animals but they only have 100 dollars. Horses cost 10 dollars cows cost 3 dollars and ducks cost 50 cents. How many animals can they buy. 10x+3y+.5z

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: A city wants to buy some animals but they only have 100 dollars. Horses cost 10 dollars cows cost 3 dollars and ducks cost 50 cents. How many animals can they buy. 10x+3y+.5z      Log On


   



Question 134750: A city wants to buy some animals but they only have 100 dollars. Horses cost 10 dollars cows cost 3 dollars and ducks cost 50 cents. How many animals can they buy.

10x+3y+.5z+100 I simplified for x and came up with x=10+-.3y+.05z now do I simplify for Y and then put both statements into the equaltion?
Please solve

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A city wants to buy some animals but they only have 100 dollars. Horses cost 10 dollars cows cost 3 dollars and ducks cost 50 cents. How many animals can they buy.
10x+3y+.5z <= 100
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Make a list of numbers of horses, cows, ducks
There are no more than 10 horses so put the numbers 0 through 10
below horses.
with zero horses list the possible numbers of cows from 0 to 33
with zero horses and zero cows you can buy 0 to 200 ducks
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With one horse the possible # of cows and ducks change.
Similarly with two horses, or three etc.
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Comment: there are a lot of possible combinations.
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Cheers,
Stan H.