Question

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?

Answer 1

There were `65` chickens and `35` pigs in the farmyard.

Explanation:

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.

Answer 2

There were `65` chickens and `35` pigs in the farmyard.

Explanation:

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.