How do you solve...........?

Question

I have to do this math worksheet for school and need help with a problem which states:

There are 13 animals in the barn. Some are chickens and some are pigs. There are 40 legs in all. How many of each animal is there?

I need to know how to setup the problem so that way i can solve it. Thanks!

Answer 1

Hi,

C + P = 13

2C + 4P = 40

Multiply the first equation by -2 and add equations.

-2(C + P = 13)

2C + 4P = 40

-2C - 2P = -26

2C + 4P = 40

----------------------

2P = 14

P = 7

C = 6

There are 7 pigs and 6 chickens in the barn. <==ANSWER

I hope that helps!! :-)

Answer 2

2chicken+4pigs=40 legs

2c+4p=40

and

c+p=13

Answer 3

Number of chickens is x, number of pigs is y.

x + y = 13

2x + 4y = 40

This is the setup; the solution is in the source section, so if you want to work it out on your own, don't look!

Answer 4

Let x = number of chickens

Let y = number of pigs

If there are 13 animals total, x + y = 13.

Since chickens have 2 legs, the number of chicken legs is 2x.

Since pigs have 4 legs, the number of pig legs is 4y.

The total number of legs is 40.

2x + 4y = 40

So now we have two equations in two unknowns.

x + y = 13 ... or x = 13 - y

Substitute 13 - y for x in

2x + 4y = 40

2 (13 - y) + 4y = 40

26 - 2y + 4y = 40

26 + 2y = 40

2y = 14

y = 7 = number of pigs. Then there are 13 - y = 6 chickens.

Check the answer:

6 chickens x 2 legs/chicken + 7 pigs x 4 legs / pig = 12 chicken legs + 28 pig legs = 40 legs