Chickens and rabbits are in the same cage, they have a total of 45 heads, 136 legs, and several chic

Solution: If there are x chickens, then there are (45-x) rabbits.

2x+4（45-x）=136

2x+180-4x=136

2x=44x=22 rabbits: 45-22=23 (only).

Hit: 22 chickens and 23 rabbits.

I'm glad to solve the above problems for you, I hope it will help you in your studies! 】≤

There are x chickens and 45-x rabbits.

2x+4（45-x)=136

2x+180-4x=136

2x=44x=22

45-x=23

A: There are 22 chickens and 23 rabbits.

Suppose the chicken has A, the rabbit has B, the chicken has 2 legs, and the rabbit has 4 legs, A+B=45

2a+4b=136

a=22 and b=23

There are x chickens and y rabbits.

x+y=45 (1)

2x+4y=136 (2)

From equation (1): y=45-x (3).

3) substitution (2) obtained.

2x+4(45-x)=136

2x+180-4x=136

2x=44x=22

y=45-22=23

Hello. If there are x chickens, then there are (45-x) rabbits, then 2x+4(45-x)=136

Just solve the equation.

I hope my answer will help you.

136-45*2）/2=23.Illustrates that there are 23 rabbits. You don't need to list any equations, you can learn all the stupid.

If there are x chickens, then there are (45-x) rabbits.

According to the title, 2x+4(45-x)=148

The solution is x=16, so there are 16 chickens and 29 rabbits.

Chickens and rabbits in the same cage. There are 45 heads and 148 legs, how many chickens and rabbits are in the cage? (Please use the equation to solve).

Solution: If there are x rabbits, then there are (45-x) chickens.

4x+2(45-x)=148

4x+90-2x=148

2x+90=148

2x=148-90

2x=58x=29

Then the chickens are: 45-29 = 16 (only).

A: There are 29 rabbits and 16 chickens.

Since there are 45 heads and 148 legs, then.

If there are x chickens in the cage, there will be 45-x rabbits

2x+4x（45-x）=148

2x-4x=148-180

2x=32x=16

Answer: There are 16 chickens and 29 rabbits.

If there are x chickens, then the rabbit will be 45-x, the chicken will have 2x legs, and the rabbit will have 4 (45-x) legs.

2x+4(45-x)=148

2x=32x=16

There are 16 chickens and 29 rabbits in the cage.

If there are x rabbits, then there are 45-x chickens.

4x+2×（45-x）=148

4x+90-2x=148

2x=148-90=58

x=2945-x=45-29=16

So 16 chickens and 29 rabbits.

Solution: If there are x chickens, then rabbits have (45-x).

According to the title, 2x+4(45-x)=148

Solution x=16

A: There are 16 chickens and 29 rabbits.

There are x chickens and y rabbits.

x+y=45

2x+4y=148

2x+2y=90

2y=58y=29

x=45-29=16

There are 16 chickens and 29 rabbits.

Set: x chickens, 45-x rabbits.

2x+(45-x)4=148

2x+180-4x=148

2x=32x=16

Rabbit = 45-16 = 29

Solution: There is a chicken and a rabbit. According to the title: 45

Solution 2 = 58

then =16

There are x chickens and y rabbits.

Let the equation x+y=45 2x+4y=148

The solution is x=16 y=29

Solution: If there are x rabbits, then there are 35-x chickens.

4x+2×（35-x）=100

4x+70-2x=100

2x=100-70

x=30÷2

x = 15 chickens: 35-15 = 20 chickens.

A: There are 15 rabbits and 20 chickens.

Solution: Suppose all 35 are chickens, then there are legs:

2 35 = 70 (strips).

The number of rabbits is: (100-70) (4-2)=15 (only) The number of chickens is: 35-15=20 (only).

A: There are 20 chickens and 15 rabbits.

Set: There are x chickens, which can be obtained from the title.

2x+（35-x）*4=100

Solve the equation to obtain: x=20

Number of rabbits = 35 - 20 = 15

So there are 20 chickens and 15 rabbits.

For a one-dimensional equation, let the chickens have x and the rabbits have (35 x).

2x+4(35－x)=100

2x=40x=20

35 20 = 15 (only).

There are 20 chickens and 15 rabbits.

Solve the general derivation of chickens and rabbits in the same cage:

Number of chicken heads = total number of heads - number of rabbit heads.

4 legs x rabbit + 2 legs x (total number of heads - number of rabbit heads) = total number of legs.

2 legs x rabbit + 2 legs x total number of heads = total number of legs.

2 legs x rabbit = total number of legs - 2 legs x total number of heads.

The general formula is obtained:

Rabbit = (total number of legs - 2 legs x total number of heads) 2Substitute the data for this question:

Rabbit = (100-2x35) 2 = 15 (only).

Chicken = 35-15 = 20 (pcs).

Let's say there are x chickens, then rabbits have (35-x) of them.

2x+4 (35-x)=100, x=20, so there are 20 chickens and 35-20=15 rabbits

So there are 20 chickens and 15 rabbits.

Solution: If there are x chickens, then there are 35-x rabbits.

2x+(35—x)*4=100

4x—2x=35*4—100

x=40/(4—2)

x=20 rabbits: 35—20=15.

Suppose there are x chickens and rabbits have y, and the equation system for this problem is: x+y=35 2x+4y=100

Note: Chickens have two legs and rabbits have four legs.

Solution: If there are x rabbits, then there are 35-x chickens. According to the title, the equation can be listed as:

4x+（35-x）×2=100

4x+70-2x=100

2x=100-70

2x=30x=15

Chickens have: 35-15=20 (only).

Answer: Slightly.

Solution: Set x rabbits and chickens (35-x).

4x+2（35-x）=100

4x+70-2x=100

2x=30x=15

35-x=35-15=20

A: There are 20 chickens and 15 rabbits.

Let there be x chickens and y rabbits, and the equation is:

x+y=35

2x+4y=100

Solution: x=20, y=15

Set x rabbits, 35-x chickens.

4x+2(35-x)=100

4x+70-2x=100

2x=100-70

x=30÷2

x = 15 rabbits 15, chickens 20 xing.

Suppose there are x chickens and y rabbits, then.

x+y= 35

2x+4y=100

Solve the equation to get x=20 y=15

So, there are 20 chickens and 15 rabbits.

Solution: If there are x rabbits, then there are 35-x chickens.

4x+2（35-x）=100

4x+70-2x=100

2x=30x=15

35-x=35-15=20

If there are x chickens and y rabbits, then there is:

x+y=35（1）

2x+4y=100（2）

2)—(1) 2.

2y=30 y=30÷2=15

x=35—15=20

So, there are 20 chickens and 15 rabbits.

Solution: Set x chickens and rabbits.

x+y=35

2x+4y=100

Solution x=20

y=15 There are 20 chickens and 15 rabbits.

Solution: Let's set the chicken as y.

y 2 (35-y) 4 100

2y 140 a 4y 100

140-100 4y-2y

40＝2yy＝20

35-20-15

A: There are 20 chickens and 15 rabbits.

Set up x chickens and rabbits.

x ten y = 35

2x ten 4y = 100

Solution x=20

y = 15 chickens 20 rabbits, 15 rabbits.

Suppose: If there are x rabbits, then there are 35-x chickens.

4x+2×（35-x）=100

4x+70-2x=100

2x=30x=15 (only).

35-15=20 (only).

A: There are 20 chickens and 15 rabbits.

feet2 chickens + 4 rabbits = 100

head: Chicken + head = 35 equal sign multiply 2, get2 chickens + 2 rabbits = 70

feetminushead(Eliminate the chickens), get 2 rabbits = 30 get free = 15 and get chickens = 20

Suppose 35 are all rabbits, then chickens have:

20 (only) rabbits: 35-20 15 (only).

Let's say there are chickens x rabbits.

x+y=35

2x+4y=100

x=20,y=15

There were twenty chickens and fifteen rabbits.

Chicken 2 legs, rabbit 4 legs, 35 2 = 70, 35 4 = 140, (100-70) (4-2) = 15, rabbit 15, (140-100) (4-2) = 20, chicken 20.

Chickens have Qinghong x only.

2*x+4*(20-x)=44

2x=36x=18

20-18=2.

Therefore, there are 18 chickens and 2 rabbits.

Solve the lack of x rabbits and 35 x chickens 4x 2 35 x 94 4x 2 35 2x 94 2x 24 x 12 35 x 35 12 21 So, there are 12 rabbits and 21 chickens. Follow-up: You seem to have miscalculated, there are 35 heads, and you count the brigade is 33 heads:

There are x rabbits and 35 x chickens 4x 2 35 x 94 4x 2 35 2x 94 2x 24 x 12 35 x 35 x 35 12 23 So, there are 12 rabbits and 23 chickens.

Solution: If there are x rabbits, then there are 35-x chickens.

4x+2×（35-x）=100

4x+70-2x=100

2x=100-70

x=30÷2

x = 15 chickens: 35-15 = 20 chickens.

Answer: There are 15 rabbits and 20 chickens.