Question 158310
This problem has been around for a looooong time.  It's yet another example of having more unknowns than equations.  In problems such as this, it's usually necessary to apply a great deal of trial and error coupled with the constraints inherent in the problem.  Many times, there is more than one solution.

Let x=number of chickens
y=number of pigs
z=number of cows

Now we are told that:
x+y+z=100----------------------------eq1
we are also told that:
0.5x+3y+10z=100------------------------eq2
Multiply eq 2 by 2 and then subtract eq1 from it and we get:

5y+19z=100
subtract 19z from each side

5y=100-19z  divide each side by 5

y=(100-19z)/5----eq2a --- this equation gives us a relationship between the number of pigs (y) and the number of cows (z).

Now we know a couple of things that will help us to solve this problem:
(1) We cannot have fractions of animals ---we are dealing with whole numbers
(2) We cannot have negative animals----we are dealing with positive numbers
(3) We must, at least, have one of each animal

By inspecting eq2a, we can quickly see that we cannot have more than 5 cows otherwise we start getting negative pigs. So, lets do some trial and error:

If we have 5 cows-----------------y=(100-95)/5=1 BINGO!!
So, 5 cows and 1 pig are possibilities
Lets substitute these values into eq1 and see if they work for chickens:
x+1+5=100
x=94 chickens
Clearly the numbers check;  let's see if the values check:
94*0.5+1*3.00+ 5*10=100
47+3+50=100
100=100
So, 94 chicks, 1 pig and 5 cows work
Let's see if we have other possibilities:
If we have 4 cows-----y=(100-76)/5----No!!! we get fractions of pigs
If we have 3 cows-----y=(100-57)/5----No!!! we get fractions of pigs
If we have 2 cows-----y=(100-38)/5----No!!! we get fractions of pigs
If we have 1 cow------y=(100-19)/5----No!!! we get fractions of pigs again
{{{0 cows would also work if the problem allowed}}

Hope this helps---ptaylor