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Let $c_{hicken} =$ Chicken & $c_{ow} =$ Cow

Since there are a total of twelve animals on the farm

$c_{hicken} + c_{ow} = 12$

Chickens have two legs and cows have four so this goes into a second equation

$2c_{hicken} + 4c_{ow} = 28$

Re-arranging the first one gives

$c_{ow} = 12 - c_{hicken}$

Substituting this into the second one makes

$2c_{hicken} + 4(12 - c_{hicken}) = 28$

This simplifies down to

$48 - 2c_{hicken} = 28$

Re-arranging this gives a value for $c_{hicken}$

$c_{hicken} = 10$

So there are ten chickens on the farm

no. of cow =x no. of chicken = y then x+y=12 4x+2y=28; 2x+y=14; x=2;y=10

let the number of chicken be x and the number of cow is 12-x the solution is : 2x+4(12-x)=28

Cows have 4 legs Chicken have 2 There are 12 animals 28 legs

To conclude: 12x2=24 so we need more 4 legs so there are 2 cows and 12-2=10 chicken.

cow+chicken=12 and we know no of legs of cow are 4 and chicken 2 so 2cow+4chicken=28 solving both equations we get 10 chickens