Question 97502
This problem involves diophantine equations, where you have more unknowns than equations.
The solutions, however, are limited by the fact that the unknowns must be integers, i.e. whole cows, whole horses, etc.
We can set up two equations as follows: H = number of horses, C = number of cows, and ch = number of chicks.

From the problem description, we can write:
H+C+ch = 100 The total number of animals is 100.
($5)H+($3)C+($0.50)ch = $100 The total amount the farmer spends.
The two equations are:

1) H+C+ch = 100
2) 5H+3C+0.5ch = 100 Multiply this equation by 2 and subtract equation 1) from the result.

1) H+C+ch = 100
2a) 10H+6C+ch = 200
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3) 9H+5C = 100 Subtract 9H from both sides.
3a) 5C = 100-9H  Now divide  both sides by 5.
3b) C = 20-(9/5)H Now C can only be an integer if H is a multiple of 5.  Let's try H = 5.
3c) C = 20-(9/5)(5)
C = 20 - 9
C = 11 This is one possibility. Now let's try H = 10
3d) C = 20 - (9/5)(10)
C = 20 - 18
C = 2 This is another possibility. 
H cannot be greater than 10 otherwise you would end up with C being a non-integer or a negative number.

So, if C = 11 when H = 5, then ch = 100-11-5 = 84 - the number of chicks.
Let's check this solution. 

H+C+ch = 100
5+11+84 = 100  and...
5($5) + 11($3) + 84($0.50) = $25 + $33 + $42 = $100 This is a valid solution!

Now if C = 2 when H = 10, then ch = 100-2-10 = 88 - the number of chicks.
Let's check this solution.
H+C+ch = 100
10+2+88 = 100 and...
10($5)+2($3)+88($0.50) = $50+$6+$44 = $100 This solution is also valid!

The two solutions are:

1) 5 horses, 11 cows, and 84 chicks.

2) 10 horses, 2 cows, and 88 chicks.