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?

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?

2 Answers

Impact of this question

16544 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-31T05:42:15

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 · 2017-01-31T05:43:06

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.

Science

Math

Humanities

... and beyond

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?

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question