# One day many years ago, while Farmer Mead was plowing her fields, she came upon an old bottle in the dirt. As she picked it up and and began to brush the dirt off, out came the Math Genie. He offered a crisp new \$100 bill to Farmer Mead, as long as she promised to spend the entire \$100, and no more, to purchase exactly 100 farm animals. She was to buy at least one cow, one sheep and one horse. When she arrived at the market, she discovered that sheep cost \$0.50 each, horses cost \$10.00 each and cows cost \$1.00 each. “Wow,” she thought, “this is going to be harder than I thought.” Can you help the farmer solve her dilemma? Find a way for Farmer Mead to spend the \$100. Be sure to explain how you solved the problem and how you know your solution is correct. Extra: Find all the possible ways Farmer Mead could spend the \$100 and satisfy the Genie's requirements. Explain how you know you have found them all. We have a total of \$100 to spend, and we must spend exactly \$100. We are to buy exactly 100 animals consisting of sheep, cows, and horses and we are required to get at least one of each. Each sheep costs \$0.50, each cow \$1 and each horse \$10.

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We have a total of \$100 to spend, and we must spend exactly \$100. We are to buy exactly 100 animals consisting of sheep, cows, and horses and we are required to get at least one of each. Each sheep costs \$0.50, each cow \$1 and each horse \$10.

Assign variables: let x equal the number of sheep purchased, y the number of cows, and z the number of horses. We can now set up a system of equations:

.5x+y+10z=100
x+y+z=100

This system has two equations with three unknowns. Thus the system is consistent (there is an answer) and dependent (there are multiple solutions.) Of course this is speaking of the mathematics. The problem has additional constraints that might make a mathematically correct solution nonviable. For instance if there are 6 horses, 108 sheep and -14 cows we have 6+108-14=100 and 6(\$10)+108(\$0.5)-14(\$1)=\$100 but we cannot have a negative number of cows.

One way to solve the system is by elimination.

I. .5x+y+10z=100
II. x+y+z=100

2(I)-II yields y+19z=100 or y=100-19z.
I-II yields -.5x+9z=0 ==> .5x=9z ==> x=18z.

Thus the solution set is {18z,100-19z,z} for `z in NN, 1<=z<=5`

For example if z=1 then x=18 and y=81. This corresponds to 1 horse, 18 sheep and 81 cows. 1+18+81=100 animals and \$10+\$9+\$81=\$100 as required.

You could solve the system in terms of any of the variables.