How do I solve this problem? There are 22 animals on a farm, with a total of 56 legs. How many chickens and...

Question

...how many hogs are there on the farm if all of the animals are either chickens or hogs?

Answer 1

x = chickens
y = hogs
chickens have 2 legs
Hogs have 4 legs.
x+y = 22
2x + 4y = 56
Now solve.
x = 22 - y
2(22-y) + 4y = 56
44 - 2y + 4y = 56
44 + 2y = 56
2y = 32
y = 16
x = 22 - 16
x = 6
Answer: 6 chickens and 16 hogs

Answer 2

If you don't know algebra (or even if you do) you can make a trial/error table. All I did was test some possible combinations until I got a possible one.

C = chickens
H = hogs

Multiply by the number of legs of each to get the total number of legs:

C(x2) H(x4) Legs

1       1       6   (not enough)
3       3       18  (not enough)
5       5       30  (not enough)
7       7       42  (not enough)
7       9       50   (not enough)

(note: I saw that I only needed 6 more here, so I added one of each animal):
8       10       56   But we need 22 animals....

16 chickens
6 hogs

I added 2 chickens and took away 1 hog until the sum was 22. (2 chickens = the same number of hog legs)

Answer 3

algebra. Let's use H to represent the number of hogs. Let C be the number of chickens. Hogs have four legs and chickens have two legs. The total number of legs is 56.

Therefore, 4*H + 2*C = 56
This can be simplified to be 2*H + C = 28

There is no single answer for this problem. If you have 10 hogs, then 8 chickens will give you 56 legs. If you have 8 hogs, then 12 chickens will give you 56 legs. If you plot this formula on a graph, it will be a line.

Answer 4

For this question, you set up a system of equations. There are two equations. First of all, you have two animals: chickens =x and hogs=y. Chicken have two legs and hogs have four legs. Therefore,
2x+4y=56.

Secondly, the total animals are 22. Therefore,

x+y=22.

You put the two equations together and get:

2x+4y=56
x+y=22

With two equations and two variables you can solve the equations as follows:

2x+4y=56
x+y=22 (multiply this equation by 2)

2x+4y=56
2x+2y=44

Subtract the two equations and get:

0+2y=12

2y=12
y=6

Then you just put the y=6 into the first or second equation to get the number of chickens which is x=16. You can check your answer by replacing the x and y values into either equation to see if you get the number on the other side (right hand side).

Answer 5

chickens = 2 legs
hogs = 4 legs

22 animals

11 hogs = 44 legs
11 chickens = 22 legs

66 legs. so that's not it.

we need 10 less legs, on the same # of animals
9 hogs = 36 legs
13 chickens = 26 legs

= 62
so..

8 hogs = 32 legs
14 chickes = 28 legs.. see the progression?
60 legs

7 hogs = 28 legs
15 chickens = 30 legs

58

6 hogs = 24 legs
16 chickens = 32 legs

56 legs


The guy directly above me is an idiot.

So is the guy directly below me.
it's not 7 chickens.

it's 6 hogs (for 24 legs) and 16 chickens (for 32 legs)

I know I did it the long way, but wanted to show you how you could "work down" to it.

Answer 6

6 hogs= 24 legs
16 chickens=32 legs
22 animals=56 legs

Answer 7

How to solve is by putting it in equation form.

H + C = 22

4*H + 2*C = 56

The first equation tells us that H = 22 -C
so we can put that in the second equation instead of H:

4*(22 - C) + 2 * C = 56

88 - 4C + 2C = 56
-2C = -32
C = 16
So, we have 16 chicken, and the other 6 animals have to be hogs.

Answer 8

the 1st answer given is right. right this is the mathematical run-down: X: style of Chickens Y: style of Hogs (a) X+Y = 22 2*X: style of chicken legs 4*Y: style of Hog Legs 2*X+4*Y = fifty six entire legs. in case you simplify by 2 on the two aspects: (b) X+2*Y = 28 From (a): X = 22-Y. Inject this into (b): X+2*Y = 22-Y+2*Y = 28. So Y = 6 and X = 22-Y = sixteen.

Answer 9

LET x be number of chickens
y be number of hogs.
x+y=22
4x+2y=56
solving these two eqns x=16
y=6

Answer 10

Hogs have 4 legs and chickens have 2, I would say whatever equals 56 legs between the two that adds up to 22 animals is your answer.

Answer 11

14 hogs