There is a flock of sheep and ducks in the meadow, with a total of 17 heads and 52 legs, how many sh

Suppose these ducks and sheep are trained to be particularly obedient and raise one foot when they hear the whistle, and after the first whistle, there are still 52-17 = 35 feet, and another blow, there are still 35-17 = 18 legs. At this time, the duck is sitting on the ground, and the sheep is standing on both feet, 18 2 = 9, then there are 9 sheep and 17-9 = 8 ducks.

Analysis: If there are x sheep, then there are 17-x ducks, multiply them by their number of legs, and add them together to 52 legs.

Column: x 4 (17 x) 2

4x＋34－2x＝52

2x 18x 9 (pcs).

17 9 8 (only).

A: There are 9 sheep and 8 ducks in the meadow.

and in the meadow there were a sheep and a bduck.

There are 17 heads, i.e.: a+b=17....

There are 52 legs, a sheep has four legs, and a duck has two legs, i.e.: 4a+2b=52....

Will - 2, 2a=18, so a=9

Substituting a=9 into , we get b=8

So there were 9 sheep and 8 ducks in the meadow.

Conventional solution: unary equations.

If there are x sheep, then there are (17-x) ducks.

Column equation: 4 times x x + 2 times (17-x) = 52 solution equation: 4x + 34-2x = 52

2x+34=52

x is equal to 9, 17-9 = 8

So there are 9 sheep and 8 ducks.

Unconventional solution: Sheep have 4 legs, ducks have 2 legs, there are 17 in total. Suppose both the sheep and the duck raise 2 legs, there are 52-17 times 2 = 18 legs left, and the ducks are sitting on the ground, and the rest of the legs are sheep's, so the sheep have 18 divided by 2 = 9, and the ducks have 17-9 = 8.

Assuming that an unknown number is known, for example, if 17 heads are all sheep, then there are a total of legs (4 17=) 68 (only), which is inconsistent with the known 52 legs in the question, and the extra (68-52=) 16 (only), the reason for the more is that each duck counts 2 more feet after being a sheep, so the number of ducks is (16 2=) 8 (only), then the number of sheep is (17-8) 9 (only).

There are 9 sheep and 8 ducks.

This belongs to the problem of chickens and rabbits in the same cage, and an unknown number x

Suppose there are x sheep, then there are (17-x) ducks.

4x+2(17-x) 52, solution x 9

So there are 9 sheep, 8 ducks. Hope!

This is the topic of the chicken and rabbit in the same cage of the primary school Olympiad.

Assuming all ducks, there are: 17*2=34 legs.

In fact, it is 52 legs, which is 52-34=18 legs less. These 18 legs are created by assuming that a four-legged sheep is a duck. Each sheep is assumed to be a duck with 4-2=2 legs missing.

18 legs means that there are 18 2=9 sheep that are assumed to be ducks. Column: Sheep:

9 (only) ducks:

8 (only) thank you.

Method 1: Sheep 4+(17-sheep) 2=52

Sheep = 9 Duck = 17-9 = 8

Method 2: Sheep + Duck = 17

4 sheep + 2 ducks = 52

Solve the equation to obtain: sheep = 9 duck = 8

Set: x sheep and y ducks.

Solution: x+y 17

4x+2y＝52

x＝9y＝8

A: There are 9 sheep and 8 ducks in the meadow.

4x sheep + 2x (17-sheep) = 52

4 sheep - 2 sheep = 52-34

Sheep = 9 (pcs).

Duck = 17-9 = 8 (only).

A: 9 sheep and 8 ducks.

Check: 4x9 + 2x8 = 52 (legs).

There are nine sheep and eight ducks.

Suppose there are x sheep and y ducks, each sheep has a head and four legs, and each duck has a head and two legs, so a system of equations can be listed.

x+y=17

4x+2y=52

9 sheep and 7 ducks were solved.

Got it?

Set sheep x, duck y

4x+2y=52

x+y=17

9 sheep and 8 ducks.

Typical chickens and rabbits in the same cage, 8 ducks, 9 sheep.

There is a flock of ducks and a flock of sheep running together, a total of 348 legs, how many sheep and how many benches are not necessarily a duck, for example, 58 and 58, 57 and 60, jujube old.

56 and 62....

There are many answers to this question. It is impossible to list them all.

Solution: It is known that the sheep has 4 Jing Xun legs and the duck has 2 legs, that is, a sheep and a duck are: 4 + 2 = 6 (pieces) The legs are a group of discussion manuscripts that are leaky and rotten, 348 6 = 58 (groups), so the sheep and ducks each have: 1x58 = 58 (only).

A: There are 58 sheep and 58 ducks.

If there are x sheep, then the duck has (10-x) and only repents (because there are several heads, there are several biju animals), because each sheep has 4 legs, each duck has 2 legs, and there are 28 legs, so: 4x+2(10-x)=28 The solution is x=4, so there are 4 sheep and 6 ducks.

27*2=54 (reputation disorder).

72-54=18 (strips).

18 (4-2) = 9 (then nothing) ......Yang Sun didn't.

27-9=18 (only) ......Duck.

With x ducks and y sheep, there is a system of equations:

x+y=10

2x+4y=24 (both sides divided by 2, x+2y=12) gives x=8, y=2, that is, there are 8 ducks and 2 sheep.

(24 feet - 2 feet x 10) (4 feet - 2 feet).

4 feet 2 feet.

2 (sheep) 10 - 2 = 8 (ducks).

A: There are 8 ducks and 2 sheep.

5 sheep and 15 ducks.

Solution: The bent finch is set up with x sheep and duck y.

x+y=20

4x+2y=50

x=5, y=15

Therefore, there are 5 early sheep and 15 spine ducks.

Think of it all as duck legs first.

Sheep = (34-12x2) 2 = 5 pcs.

Ducks = 12-5 = 7 pcs.

Seven ducks, five sheep, it's a very simple operation.

Suppose there are x sheep and 8-x ducks.

2×（8-x）+4×x=26

16-2x+4x=26

2x=10x=58-5=3

3 ducks and 5 sheep.