Each chicken has 2 legs and each cow has 4 legs. The farmer has a total of 84 animals (chickens + cows) and a total of 218 legs. Let:
- ccc = number of chickens
- www = number of cows
We have two equations based on the problem:
- c+w=84c+w=84c+w=84 (total animals)
- 2c+4w=2182c+4w=2182c+4w=218 (total legs)
From the first equation, express www as:
w=84−cw=84-cw=84−c Substitute into the second equation:
2c+4(84−c)=2182c+4(84-c)=2182c+4(84−c)=218 Simplify:
2c+336−4c=2182c+336-4c=2182c+336−4c=218
−2c+336=218-2c+336=218−2c+336=218
−2c=218−336-2c=218-336−2c=218−336
−2c=−118-2c=-118−2c=−118
c=−118−2=59c=\frac{-118}{-2}=59c=−2−118=59 So, the farmer has 59 chickens
and substituting c=59c=59c=59 into the first equation gives cows:
w=84−59=25w=84-59=25w=84−59=25 Thus, the farmer has 59 chickens.
