There are 100 chickens and rabbits, and the number of chicken feet is 28 less than the number of rab

The number of chicken feet is 28 less than the number of rabbit feet, and if you add 14 chickens, the chicken feet and rabbit feet are the same.

With the same feet, the number of heads of a chicken is twice as large as that of a rabbit.

114 (2+1)=38pcs.

The number of chickens is 38*2-14=62.

The number of rabbits is 38*1=38.

There are x chickens and y rabbits.

x + y =100

2x=4y-28

Result: x=100-y

Substituting into , we get 2(100-y)=4y-28 and we get y=38 into y=, and we get x=100-38=

There are 68 chickens and 38 rabbits.

114 (2+1)=38pcs.

The number of chickens is 38*2-14=62.

The number of rabbits is 38*1=38.

Elementary school topics only do this, xy's did not learn.

There are x chickens and 100-x rabbits.

Then the chicken feet plus the rabbit feet have 2x+28=4(100-x). Calculate x=62. 38 rabbits.

Set x chickens and y rabbits.

x + y =100

2x = 4y-28

The solution is x=62 , y=38

So there are 38 rabbits and 62 chickens.

Set x chickens and y rabbits.

x + y =100

2x = 4y-28

The solution is x=62 , y=38

It must be right!

Solution: Set x chickens and y rabbits.

x + y =100 ①

2x=4y-28 ②

Obtained by 1: x=100-y

Substituting 3 into 2 to get 2 (100-y) = 4y-28 solution is y = 38 and substituting y=38 into 1 to get x = 100-38 = 68 This is the step of solving the problem in junior high school.

There are x chickens and y rabbits.

4x+2y=100

2y-4x=28

Get x=8 y=27

So there are 8 chickens and 27 rabbits.

This is a typical Olympiad problem for elementary school students, and it is too easy. You don't want to ask people everything, you have to ask yourself first if you have a problem, understand? Only then will you be successful.

There are x chickens and y rabbits.

4x+2y=100

2y-4x=28

Get x=8 y=27

x + y =100 ①

2x=4y-28 ②

Obtained by 1: x=100-y

Substituting 3 into 2 gives 2(100-y)=4y-28 and substituting y=38 into 1 to get x=100-38=68

All of the above are true, 62 chickens and 38 rabbits.

You can go to death - -

What era is this, if you don't blush, I'll blush for you!

Assuming that all rabbits are rabbits, there should be 100*4=400 more rabbit feet than chicken feet, and 400-28=372 more than the actual ones.

Replace a rabbit with a chicken, and the difference in feet will be 4+2=6.

Number of chickens = 372 6 = 62.

Number of rabbits = 100-62 = 38.

Solution: If there are x chickens, then there are (100-x) rabbits, and the equation is listed according to the problem

4（100-x）-2x=28

400-4x-2x=28

400-6x=28

6x=372

x = 62 rabbits: 100-x 100-62 38 (only).

Suppose there are x chickens and y rabbits, and list two equations according to the title: x+y=100 2*x-4*y=20

The two equations are solved to get two unknowns, x=70, y=30, i.e. there are 70 chickens and 30 rabbits.

Solution: If there are x chickens, then there are (100-x) rabbits, the total number of chickens is 2x, and the total number of rabbits is 4 (100-x), 2x-4(100-x) = 20

Solve this equation: 2x-400+4x=20

6x=420

x=70 rabbits have 100-70=30.

A: There are 70 chickens and 30 rabbits.

Set x chickens and rabbits (100-x).

2x-4(100-x)=80

6x=480

x = 80 chickens and 20 rabbits.

114 (2+4)=19pcs.

Rabbits: 19 2 = 38 pcs.

Chickens: 100-28 = 62 chickens.

Let the rabbit be the x-column equation 4x-2(100-x)=28 x=38

So rabbit 38 chicken 62

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

2（100-x）-4x=20

200-2x-4x=20

200-6x=20

6x=200-20

6x=180

x=180÷6

x=30100-30=70 (only) ......Chicken.

20 more feet is ten chickens, 100 minus 10 chickens, and 90 are left. The chicken has two legs, the rabbit has four legs, that is, the chicken accounts for two-thirds of the 90, the rabbit accounts for one-third, the chicken is 60, the rabbit is 30, the chicken plus the ten is 70 chickens, and the rabbit is 30.

There are 200 chickens and rabbits, and the number of chickens and rabbits is 160 more than the number of rabbits. How many chickens and rabbits are there?

Chicken (4 200 + 160) (4 + 2).

160 pcs. Rabbits 200-160 = 40.

I didn't learn equations in the third grade, so I used the knowledge of the third grade.

Suppose 100 are all chickens, and legs 100 2 = 200 (only) 200 + 28 = 228 (only).

228 (4+2)=38 (number of rabbits) 100-38=62 (number of chickens).

Set the chicken is x pcs.

Rabbits are 100-x

2x+28=4（100-x）

2x+28=400-4x

6x=400-28

x=372/6

x=62.

Rabbits are 100-62=38 pcs.

If there are x chickens, then 100-x rabbits.

2x+28=4（100-x）x=62

A: There are 62 chickens and 38 rabbits.

Suppose there are x chickens, then there are 100-x rabbits, and we get equation 4 (100-x) = 2x + 28

The solution is x=62,100-x=38

So there are 62 chickens and 38 rabbits.

Suppose chickens have x and rabbits have 100-x 2x+28=4(100-x).

6x=372

x=62, so there are 62 chickens and 38 rabbits.

Set: x chickens and y ducks.

x+y=100

4y-2x=28

I don't want to forget it, I don't have a pen.

Chickens have 62 124 legs, rabbits 38 have 152 legs, and chickens have 28 feet less than rabbits.

If there are 50 chickens and 50 rabbits, then 50 chickens are less than rabbits*(4-2)=100 feet, and for every rabbit decreased, the difference in the number of legs between rabbits and chickens decreases by 4+2=6(100-28) 6=12.

So to reduce 12 rabbits and add 12 chickens, so the number of rabbits is 50-12 = 38.

The number of chickens is 50 + 12 = 62.

There are 62 chickens and 38 rabbits.

Solution: There are x chickens and y rabbits.

Then the system of equations that can be listed according to the meaning of the question is, x+y=100

2x-4y=-28 ②

From the formula we get x=100-y, substituting x=100-y into the formula we get, 200-2y-4y=-28, we get y=38, substituting y=38 into the formula, we get x=62, then x=62, y=38.

That is, there are 62 chickens and 38 rabbits.

Solution 1: If you add 28 chicken feet, that is, there are 28 2 = 14 chickens, the number of chickens and rabbit feet is equal, and the rabbit's feet are the chicken's feet 4 2 = 2 (times), so the number of chickens is 2 times the number of rabbits. The number of rabbits is.

100+28 2) (2+1)=38 (only)

Chickens are 100-38 = 62 (only).

A: 62 chickens and 38 rabbits.

Of course, you can also remove rabbit 28 4 = 7 (only).

The number of rabbits is.

100-28 4) (2 + 1) + 7 = 38 (only).

It is also possible to assume an arbitrary number.

Solution 2: Suppose there are 50 chickens, there are rabbits 100-50=50 (only).At this time, the difference in the number of legs is 4 50-2 50 = 100, which is 72 more than 28It means that the hypothetical number of rabbits is more (the number of chickens is less).In order to keep the total number of 100, one rabbit was replaced by a chicken, with 4 less rabbit feet and 2 more chicken feet, a difference of 6 (don't be careful, not 2).So the number of rabbits to be reduced is:

100-28) (4+2)=12 (only)

The number of rabbits is. 50-12=38 (only)

Then the number of chickens can be obtained naturally.

100 38 = 62 (only).

Adding it to the first formula, we get: 3y=174, that is, y=58 can be substituted into the first formula, x=42

Therefore, there are 42 chickens and 58 rabbits.

Also, looked at the answers above.

The rabbit is four-legged, I guess I watched Bugs Bunny too much, at first I thought it was two legs, I naturally didn't learn well in elementary school, hehe.

Finally, I wish you a happy New Year's Day!

Got it, revise it again.

The second formula becomes 4y-x=148

The addition of the two formulas yields 5*y=248 y=

Divisible, obviously not true...

In this case, there is a possibility for the following.

1.There are one or several rabbits or chickens that are slightly disabled.

2.There are n rabbits that are standing, or there are m chickens that are independent in the golden rooster.

The chicken has x, and the rabbit has y

Then, we can know: x+y=100

4y-2x=28

Then calculate this system of binary linear equations.

38 rabbits were solved.

62 chickens.

Solution: Let there be x rabbits and (100-x) chickens.

2(100-x)+28=4x

228-2x=4x

6x=228

x = 38 chickens: 100-38 = 62 (chickens).

A: There are 38 rabbits and 62 chickens.

Let the chickens x and the rabbits be y, then x+y=100, 4y-2x=28, and solve the equation x=62, y=38

Chicken 43, rabbit 57

28 divided by 2 equals 14, which means that there are 14 fewer chickens than rabbits.

100 divided by 2 = 50 14 2 = 7 chickens are 50-7 = 43 rabbits are 50 + 7 = 57 , so they differ by 14.

Equation: 28 2 2=7 100 2=50 rabbits: 50+7=57 chickens: 50-7=43.

62 chickens and 38 rabbits.

The chicken has x, and the rabbit has y

Then, we can know: x+y=100

4y-2x=28

Then calculate this system of binary linear equations.

38 rabbits were solved.

62 chickens.

Suppose there are x rabbits and y chickens. List the system of equations x+y=100 4x-2y=28 and solve x=38 y=62 so there are 38 rabbits and 62 chickens.

There are x chickens.

There are 100-x rabbits.

Rabbits have 40 more feet than chickens

(100-x)*4=2x+40

x=60.

Chicken 60, rabbit 40

Set x chickens and y rabbits. The number of chicken legs is 2x, and the number of rabbit legs is 4y. x+y=100, 2x=4y-28, and the solution is bidynamic.

First, set the chicken as x and the rabbit as the rabbit.

4y-2x=28

In the end, there were 62 chickens and 38 rabbits.

Let the chicken x, the rabbit y, x+y=100

4y-2x=28

Solve the equation to get x=62,y=38

So, 62 chickens and 38 rabbits.

Set chicken A and rabbit B.

then a+b=100

4b-2a=28

The solution is a=62 and b=38

So, 62 chickens and 38 rabbits.

There are x chickens and rabbits have y.

x+y=100

4y-2x=28

The solution is x=62 , y=38

Can't do this question? Actually, I won't!

Count 28 feet first, which is equivalent to the number of rabbits 28 4 = 7 (only) and a few chickens and rabbits left 100-7 = 93 (only).

A chicken and a rabbit have a total of several legs 4 + 2 = 6 (pieces) The remaining 93 chickens and rabbits have the same number of legs, and the number of chickens accounts for 4 6, 93 4 6 = 62 (pieces).

The total number of rabbits is 100-62=38 (only).

This problem can also be solved using a one-dimensional equation.

If there are x chickens, then the rabbits have (100-x) so there is the equation: 4(100-x) -2x=28 and the solution is: x=62

That is: chickens: 62, rabbits: 38.

Hypothetical solution:

Assuming that all rabbits are 4 100 = 400 feet more than chickens, if you retreat a rabbit and add a chicken, the difference is 4 + 2 = 6 feet, there are chickens (400-28) (4 + 2) = 62 (only) and 100-62 = 38 (only) rabbits