An algebra teacher drove by a farmyard full of chickens and pigs. The teacher happened to notice that there were a total of 100 heads and 270 legs. How many chickens were there? How many pigs were there?

We shall represent the chickens as #x# and the pigs as #y#. From the data of heads and feet, we can write two equations:

#x+y=100#, since each have one head.
#2x+4y=270#, since chickens have two legs and pigs have four.

From the first equation we can determine a value for #x#.

#x+y=100#

Subtract #y# from each side.

#x=100-y#

In the second equation, substitute #x# with #color(red)((100-y))#.

#2x+4y=270#

#2color(red)((100-y))+4y=270#

Open the brackets and simplify.

#200-2y+4y=270#

#200+2y=270#

Subtract #200# from each side.

#2y=70#

Divide both sides by #2#.

#y=35#, the number of pigs.

In the first equation, substitute #y# with #color(blue)(35)#.

#x+y=100#

#x+color(blue)(35)=100#

Subtract #35# from each side.

#x=65#, the number of chickens.

We shall represent the chickens as #x# and the pigs as #y#. From the data of heads and feet, we can write two equations:

#x+y=100#, since each have one head.
#2x+4y=270#, since chickens have two legs and pigs have four.

From the first equation we can determine a value for #x#.

#x+y=100#

Subtract #y# from each side.

#x=100-y#

In the second equation, substitute #x# with #color(red)((100-y))#.

#2x+4y=270#

#2color(red)((100-y))+4y=270#

Open the brackets and simplify.

#200-2y+4y=270#

#200+2y=270#

Subtract #200# from each side.

#2y=70#

Divide both sides by #2#.

#y=35#, the number of pigs.

In the first equation, substitute #y# with #color(blue)(35)#.

#x+y=100#

#x+color(blue)(35)=100#

Subtract #35# from each side.

#x=65#, the number of chickens.