Chickens and rabbits in the same cage, a total of 15, the number of rabbit feet than the number of c

There are 10 rabbits and 5 chickens. There are two ways to solve the problem.

The first method: Let the rabbit have x and the chicken have y, and the problem can get x+y=15, 4x-2y=30. It can be solved with x=10 and y=5.

The second method: Suppose all 15 rabbits are rabbits, then the rabbits will have 15 more 4 = 60 feet than the chickens. Now there are only 30 more legs, and if the difference between the number and the actual number is (60-30) = 30 feet, then it is impossible for all of them to be rabbits, and some of them are chickens.

For each rabbit replaced with a chicken, the difference in the number of legs varies by 4 + 2 = 6. (60-30) 6=5 is the number of chickens, so if there are 5 chickens, then 10 rabbits.

If the rabbit has 30 more feet than the chicken, then the rabbit has 30 more than the chicken 2 = 15 That's 0 chickens and 15 rabbits.

Let the chickens be x and the rabbits be y, x+y=15

4y-2x=30

x=5,y=10

There are 5 chickens and 10 rabbits.

Let the rabbit x only, then.

4x-2(15-x)=30

6x=30+30=60, x=10 (only).

Chickens: 15-10 = 5 (only).

There are x rabbits and 15-x chickens

4x—2（15—x）=30

x=10，15—10=5

There are 10 rabbits and 5 chickens.

The feet of the chickens are 60 more than the feet of the rabbit, i.e. 60 2 30 chickens, 45-30 15 (the feet of the remaining chickens are equal to the feet of the rabbit, where the chickens are twice as coarse as the rabbit's stool).

15 (1+2) 5 (rabbit).

45-5 40 (chickens).

Solution: Suppose there are x chickens and (x-20) rabbits.

Because there are 20 fewer rabbits than chickens, and chickens have two legs, the equivalence relationship between rabbit four feet is chicken * foot + rabbit * foot = 262

From the meaning of the title, we can get 2x+4(x-20)=262

2x+4x-80=262

6x=340

Solution x=57

So 57 chickens and 37 rabbits.

I hope it can help you, if you have any questions, please ask in the follow-up, thank you!

If there are x chickens, then there are x-20 rabbits.

Easy to obtain: 2x+4(x-20)=262

After simple arithmetic, x=57 x-20=37

So 57 chickens and 37 rabbits.

Suppose there are x rabbits and x + 20 chickens.

then 4x+2*(x+20)=262

6x=222

x = 37 rabbits and 37 chickens 37 + 20 = 57.

Lots of chickens'20 yes, now let's take out these chickens, there are 40 legs missing, 222 chickens and rabbits are left, and now each chicken is tied to a rabbit on its back, and the combination is two with six legs, 222 is 37 except 6, so there are 37 heads and 2 are not like, there are 37 rabbits, there are 57 chickens.

If the rabbit has x, then it has 4x legs, and the chicken has x+20, then it has legs (x+20) multiplied by 2 legs, and 4x+2x+40 equals 262 to get x is 37, so the rabbit has 37 and the chicken has 57

20*2=40

After subtracting the extra 20 chickens, the chickens and rabbits have a total of 262-40 = 222 feet 222 6 = 37

There are 37 + 20 = 57 chickens.

There are 37 in the picture.

Before doing this, it is important to know that there is a hidden condition: each chicken has 2 legs, and each rabbit has 4 legs.

Solution: If there are x chickens, then rabbits have (100-x).

4(100-x)-2x=70

400-4x-2x=70

4x-2x=70-400

6x=-330

x=55 (only).

A: There are 55 chickens and 45 rabbits.

The extra 15 chickens with legs: 15 2 = 30.

The number of chickens and rabbits left is the same, there are a total of feet: 282-30=252 The remaining chickens and rabbits have each: 252 (2+4)=42 Originally, the rabbits have: 42.

There are 42 + 15 = 57 chickens.

There are 28 chickens and 43 rabbits.

Analysis: Set: There are x chickens and y rabbits.

From the problem, we can see that y-x=15 2x+4y=228 solves: x=28 y=43

So there are 28 chickens and 43 rabbits.

Chickens and rabbits in the same cage, a total of 15, the number of rabbit feet than the number of chickens 30 more, how many chickens and rabbits

Solution: There are x chickens and x - 10 rabbits.

2x + 4×(x - 10)= 260

6x - 40 = 260

6x = 300

x = 50

x - 10 = 50 - 10 = 40 A: 50 chickens and 40 rabbits.

There are 10 more chickens than rabbits.

Set Chicken x Rabbit X-10

2x+4(x-10)=260

2x+4x-40=260

6x=300

x=50 rabbits 60-10=50.

Chickens and rabbits have a total of 100 feet in the same cage, and the number of rabbit feet is 40 more than the number of chicken feet, then there are 70 rabbit feet, 30 chicken feet, and rabbit feet? Obviously wrong. It should be:

There are a total of 100 chickens and rabbits in the same cage, and the number of rabbits' feet is 40 more than the number of chickens' feet, and there are x rabbits and 100-x chickens.

4x=2(100-x)+40, the solution is x=40, and 100-x=60

4x+2y=100

4x-2y=40

So x=140 8=35 2=

The number of rabbits is not an integer, so there is no solution to this problem.