C = cows
P = pigs
K = chickens

10C + 3P + 0.5K = 100, eq#1

C + P + K = 100, eq#2

multiply eq#1 by 2
20C + 6P + K = 200

subtract eq#2
19C + 5P = 100

divide by 5
19C/5 + P = 20

C cannot be more than 5

if C = 5
P = 1
K = 94

if C = 0
P = 20
K = 80

This is a very tricky question because we have 2 equations, but 3 unknowns.

100=10C+3P+0.50K
C+P+K=100

C=cows
P=pigs
K=chickens

We know that there cannot be very many cows, so let’s try this with only pigs and chickens, factoring in the cows.

1 cow:
90=3P+0.50K
P+K=99
This is because if we account for one cow, the cost of the pigs and chickens will be $90 and there will be 99 chickens and pigs.
In this, we get 16.2 pigs. This cannot be correct, so we can try 2 cows.

2 cows:
80=3P+0.50K
P+K=98
Here, we get 12.4 pigs. This cannot be correct, so we can try 3 cows.

3 cows:
70=3P+0.50K
P+K=97
Here we get 8.8 pigs. This cannot be correct, so we can try 4 cows.

4 cows:
60=3P+0.50K
96=P+K
Here we get 4.8 pigs.

5 cows:
50=3P+0.50K
95=P+K
Here we get one pig, 94 chickens, and 5 cows.

I hope this helps.