SOLUTION: A farmer has chickens and goats,with at least one of each. The chickens and goats together have a total of 100 legs. (4 legs per goat and 2 legs per chicken) How many different com

You are letting x be the number of goats and y be the number 
of chickens.  Divide that through by 2

   2x+y=50
      y=50-2x

If they were all goats and no chickens,
thre would be x = 100/4 = 25 goats. But we must
have at least 1 chicken, so the number of
goats, x, can only be 1 to 24, inclusive. So 
there are 24 combinations of goats and chickens.

 1.   x=1 goat  and y=50-2(1)  = 50-2  = 48 chickens.
 2.   x=2 goats and y=50-2(2)  = 50-4  = 46 chickens.
 3.   x=3 goats and y=50-2(3)  = 50-6  = 44 chickens.
 4.   x=4 goats and y=50-2(4)  = 50-8  = 42 chickens.
 5.   x=5 goats and y=50-2(5)  = 50-10 = 40 chickens.
 6.   x=6 goats and y=50-2(6)  = 50-12 = 38 chickens.
 7.   x=7 goats and y=50-2(7)  = 50-14 = 36 chickens.
 8.   x=8 goats and y=50-2(8)  = 50-16 = 34 chickens.
 9.   x=9 goats and y=50-2(9)  = 50-18 = 32 chickens.
10.  x=10 goats and y=50-2(10) = 50-20 = 30 chickens.
11.  x=11 goats and y=50-2(11) = 50-22 = 28 chickens.
12.  x=12 goats and y=50-2(12) = 50-24 = 26 chickens.
13.  x=13 goats and y=50-2(13) = 50-26 = 24 chickens.
14.  x=14 goats and y=50-2(14) = 50-28 = 22 chickens.
15.  x=15 goats and y=50-2(15) = 50-30 = 20 chickens.
16.  x=16 goats and y=50-2(16) = 50-32 = 18 chickens.
17.  x=17 goats and y=50-2(17) = 50-34 = 16 chickens.
18.  x=18 goats and y=50-2(18) = 50-36 = 14 chickens.
19.  x=19 goats and y=50-2(19) = 50-38 = 12 chickens.
20.  x=20 goats and y=50-2(20) = 50-40 = 10 chickens.
21.  x=21 goats and y=50-2(21) = 50-42 = 8  chickens.
22.  x=22 goats and y=50-2(22) = 50-44 = 6  chickens.
23.  x=23 goats and y=50-2(23) = 50-46 = 4  chickens.
24.  x=24 goats and y=50-2(24) = 50-48 = 2  chickens.

Edwin