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
combinations of chickens and goats could there be? Sorry
for not knowing the right category. But all I have so far
is 4x+2y=100
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