Ten points to the best answerer: easy question, harder twist!?

Question

There are a total of 20 chickens and cows in a barnyard.
The animals have 52 legs in all. How many of each animal are there? Also, show how to use linear combinations to describe the situation. Find the correct solution.

Answer 1

Let x be the number of chickens.

Hence, 20 - x is the number of cows.

Number of legs:

2x + 4(20-x) = 52

-2x + 80 = 52

2x = 28

x = 14 Chickens

20 - 14 = 6 Cows

Bozo

Answer 2

20 chicken + x cows = 52 legs
20 * 2 + x * 4 = 52
40 + 4x = 52
4x = 52 - 40 = 12
4x = 12
x = 12/4 = 3

3 cows.

Answer 3

let x = number of chickens
let y = number of cows

so x + y = 20

now let 2x = number of chicken legs
and let 4y = number of cow legs

so 2x + 4y = 52

now solve the first equation for x to get x = 20-y.

substitute this x into the second equation to get 2(20-y) + 4y = 52

Solve for y
40 -2y + 4y = 52
y = 6 So there are 6 cows.

substitute this y into either of the first two equations (use the first)

x + 6 = 20
x = 14

so there are 14 chickens.

6 cows and 14 chickens, solved using simultaneous linear equations.

Answer 4

14 chickens and 6 cows

redhed

Answer 5

ditto for what he said