Let x = # of horses

    y = # of cows

    z = # of chicks


Then

     x  +  y +    z = 100     (1)

     5x + 3y + 0.5z = 100     (2)


Multiply equation (2) by 2 (both sides).  Keep equation (1) as is.

     x  +  y +   z = 100      (3)

    10x + 6y +   z = 100      (4)


Subtract equation (3) from equation (4).  You will get

     9x + 5y       = 100      (5)


All equations (1) - (5) are in integer numbers x, y, z. In particular, equation (5) is in integer numbers x and y.    


In equation (5), right side and the term 5y are multiples of 5.  Hence, the term 9x in equation (5) is a multiple of 5.


It gives for x only these two possibilities:  x= 5  or  x= 10.

So, consider both cases separately.


    a)  If  x= 5,  then from (5)  5y = 100 - 9*5 = 55;  hence,  y= 11.

                   Then from (1)  z = 100 - 5 - 11 = 84.


    b)  If  x= 10,  then from (5)  5y = 100 - 9*10 = 10;  hence,  y= 2.

                   Then from (1)  z = 100 - 10 - 2 = 88.


Thus the problem has two solutions in integer numbers:

    1)  5 horses,  11 cows  and  84 chicks,

and

    2)  10 horses,  2 cows  and  88 chicks.