Chickens and rabbits are in the same cage, with a total of 20 heads and 30 legs, how many chickens a

Chickens and rabbits are in the same cage, with a total of 20 heads and 30 legs, which is impossible, and if they are both chickens, they should also have 40 legs, and it cannot be 30 legs. If there are 50 legs, chickens have (20 4-50) 2 = 15 and rabbits have 20-15 = 5.

There is no solution to this question. Because:

A chicken has a head with two legs, and a rabbit has a head with four legs.

There are 20 heads in the cage, and even if they are all chickens, they have at least 40 legs. But there are only 30 legs in this question.

Turn. If there are only 30 legs, there can only be a maximum of 15 heads in the cage.

So there is no solution to this problem.

The chicken-rabbit problem is a traditional math problem. Your question is wrong with the numbers, let's say 20 chickens should have 40 legs. If it's 20 rabbits, it should have 80 legs.

Chicken and rabbit mix, the number of legs should be less than 80 and more than 40. Suppose it is 50 legs, and the calculation method is: if all 20 heads are chickens, then there are 40 legs, on this basis, there is one rabbit, then there are two more legs, and now 50 is 10 more than 40, so there are 5 rabbits and 15 chickens.

This problem can be calculated using a one-dimensional equation.

With x chickens, there are 20-x rabbits.

2x+4(20-x)=30

2x+80-4x=30

2x=50x=25

30-25=5.

A: There are 25 chickens and 5 rabbits.

The title is wrong, if it is only a chicken, 20 is 40 legs, and it is impossible to have a total of 30 legs with rabbits.

Let's say there are x chickens and y rabbits. then x+y=20;2x+4y=30；There is no solution to this question. Even if the cage is full of chickens, a chicken has a head and two legs, then 20 must have 40 legs, not to mention that rabbits have 4 legs, and there are at least more than 40 legs in the cage.

There is a problem with this question. 20 heads, indicating that the total number of chickens and rabbits is 20. If it's all chickens, there are also 40 legs. Contradictory with a total of 30 legs. Therefore, this question cannot be established.

20 heads only have 30 legs, and one head can't fit two legs on average, this is a group of monsters!

There is no solution to the problem.

Chickens have 2 legs, rabbits have 4 legs, and even 20 chickens have at least 40 legs.

This chicken and rabbit have broken hands and feet, even if there are 20 chickens, there are more than 30 legs.

You have a problem with this question, assuming that it is all chickens and 20 heads, then what are 40 legs. Where did those 10 legs go?

10 2 5 (only).

5 4 20 (strips).

30-20 10 (strips).

10 2 5 (only).

A: There are 5 rabbits and 5 chickens.

If 20 of them are chickens, they also have 40 legs...

Do you have a one-legged chicken and a two-legged rabbit?

Assuming there are x chickens and y rabbits in the cage, the number of their legs is 4x + 4y (since each chicken has 2 legs and each rabbit also has 2 legs) and the number of their heads is x + y.

The number of legs in the known cage in the problem is 36 and the number of heads is 14. So the following two equations can be listed:

4x + 4y = 36 (Kano's equation 1).

x + y = 14 (Equation 2).

By simplifying Equation 1, we get x + y = 9.

Substituting this equation into Equation 2 yields:

x + 9 - x) =14

Solving the equation yields x = 5

Because of this, there were 5 chickens and 9 rabbits in the cage.

Answer: There are 5 chickens and 9 rabbits.

It can be seen from the celebration question that chickens and rabbits have 14 heads and 36 legs, and it is known that chickens and rabbits have only one head, chickens have two legs, and rabbits have four legs.

Chicken + Rabbit = 14

2 chickens + 4 rabbits = 36

Multiply the first equation by 4 and subtract the second equation to find chickens = 10.

Bringing chickens into any of the above formulas can find rabbits = 4 filial piety clans.

According to the calculation, there are ten chickens, and there are four rabbits, here, we can set the number of chickens as y, the number of rabbits as x, and a system of binary equations to solve the collapse, of course, you can also set the number of chickens as x, then the number of rabbits is 14-x, and the unary equation is 2x+4(14-x)=56, and the number of chickens and rabbits can also be obtained by solving this equation.

There are chickens and rabbits in the cage dust dig, a total of 36 legs, how many chickens are 14 heads? How many rabbits.

Set up 14 all rabbit mill halls, then the chicken has a blind brother hidden:

14x4-36）÷（4-2）

10 (only) rabbits have:

4 (only).

Suppose it's all chickens, then there are legs:

25x2=50 bars.

70 50 = 20 rabbits: 20 (4 2)=10.

There are chickens: 25 10=15.

The chicken has 2 legs and the rabbit Kailing has 4 legs.

Let the chicken be x and the rabbit be y

2x+4y=28

x+y=10

The solution is equal to 6 y, which is equal to 4

There are 6 chickens and 4 rabbits.

If there are x chickens, then there are 10-x rabbits.

2x+4(10-x)=32

2x=-8x=4

There are 4 chickens and 6 rabbits.

15 rabbits and 5 chickens: 70 2 = 35, so there are 35 pairs of legs, chickens have only 1 pair of legs, rabbits have 2 pairs, and chickens and rabbits have 20 together. 35-20=15, so 15 rabbits and 5 chickens.

Set: chicken x, rabbit y.

2x+4y=70

x+y=20

x=20-y

2(20-y)+4y=70

40+2y=70

y=15,x=5

5 chickens and 15 rabbits.

Suppose they are all rabbits, 10x4=40 legs, now there are only 28 legs, 40-28=12, it means that there are 6 animals with 2 legs, 10-6=4, 4 rabbits, 6 chickens.

Six chickens and four rabbits.

Do you want the process. Let the chicken be x and the rabbit be 10-x;

2x+4*（10-x）=28；The solution is x=6;