cows and turkeys

let x be number of cows and y be number of turkey. therefore as per problem x+y=20 and 4x+2y=56 by solving in simultaneous equation. we get cows =8

My solution is same as @John Claude Ceriola , but I am going to explain it in detail.

Let all of the cows lift two of their legs up, and let the turkeys do a foot stand,

then the number of legs on the ground become $56/2=28$ ,

Next, let the cows do a foot stand, and let the turkeys fly into the blue sky,

then the number of legs on the ground become $28-20=8$ .

Now, every cow stands on the ground with only a leg,

therefore there are $\boxed8$ cows, and $20-8=12$ turkeys.

Heres a trick: (number of legs):56 Divide it by 2 so 28. (number of animals):20 Subtract 28-20 so you get 8 which is always the number of cows.