Chickens and rabbits in the same cage problem column equation! Please, everybody!

1. There are some red balloons and yellow balloons in a store, the number of red balloons is only 4 times that of yellow balloons, 2 red balloons and a yellow balloon are sold every day, and after a few days, there are 12 red balloons left, and the yellow balloons are just sold out, how many red balloons are there?

Solution: There were x yellow balloons, and they got to:

4x-2x=12

x=6, so it turns out that there are 24 red balloons.

2. A batch of steel, 30 cars will be used to load it at one time, 35 cars will be used to load it at one time, and each car will be loaded 3 tons less than each car, how many tons of steel are there in this batch?

Solution: If there are x tons of steel in this batch of steel, you get:

x/30-x/35=3

5x=3150

x=630, so this batch of steel has 630 tons.

3. A workshop produces 250 pieces of clothing, and can get 25 yuan for the production of a piece, if one does not meet the requirements, it will be deducted 20 yuan, and a total of 5350 yuan will be obtained after production, how many pieces are unqualified?

Solution: Set x pieces to be unqualified, get:

25（250-x）-20x=5350

45x=900

x=20 so 20 pieces are not qualified.

4. A company delivered relief grain to the disaster area in Sichuan, a total of 1,000 bags of rice and flour, 25kg of rice per package, 15kg of flour per package, 1 ton more rice than flour, how many tons of this batch of grain?

Solution: Let the flour have x packs, and get:

25（1000-x）-15x=1000

40x=24000

x=600 then there are 400 packs of rice.

So this batch of grain is shared.

400 25 + 15 600) 1000 = 19 tons.

5. There is a group of classmates on the playground, the number of boys is 4 times that of girls, each time there are 2 boys and 1 girl back to the classroom at the same time, after a few times, there are 8 boys left and 1 girl left. How many students are there on the playground?

Solution: There are x girls on the playground, and they have to:

4x-2（x-1）=8

2x=6x=3

So there are 3 4 = 12 boys on the playground.

So there were 15 students on the playground.

6. There is a bunch of black and white chess pieces, of which the number of black chess pieces is 2 times that of white chess pieces, if you take out 4 black chess pieces and 3 white flags from this pile of chess pieces at the same time, then how many times after taking it, there may be 1 white flag left, and 18 black chess pieces left?

Solution: Set up x times, get:

2（1+3x）=18+4x

2x=16x=8

So after taking 8 times, there is a situation where there is 1 white flag left and 18 black pieces left.

7. Shipped a batch of watermelons, ready to be divided into two categories, large per kilogram yuan, small per kilogram yuan, so that this batch of watermelons can be sold for a total of 290 yuan, if the size of the watermelon per kilogram of price reduction yuan, then this batch of watermelons can only be sold for 250 yuan. How many kilograms of large watermelons are there?

Solution: Equipped with x kilograms of large watermelons, get:

x=500 so there are 500 kg of large watermelons.

1.There were x red balloons.

x-x/2-12=0

x=242.A total of x tons of steel is set.

x/30-x/35=3

x=6303.There are x pieces of unqualified.

250-x)*25-20x=5350

x=204.Set up rice x packets and flour y packets.

x+y=1000

25x-15y=1000

x=400y=600

25x+15y=19000（kg）

5.There are x number of boys.

x-2*(x/4-1)=8

x = 12 for a total of 12 + 3 = 15 people.

6.Set x black chess pieces, take y times.

x-4y=18

x/2-3y=1

x=50y=8

7.Set large watermelon x kilograms, small watermelon y kilograms.

x=500y=300

x+y=800

Ask me again if you have any questions.

Chickens and rabbits in the same cage with the equation grinding source calendar split circle to solve is relatively simple, calculation method:

1. If there are x chickens, y rabbits, and 35 chickens and rabbits in the same cage, then x+y=35.

2. Chickens and rabbits have a total of 94 legs, chickens have 2 feet, and rabbits have 4 legs, then 2x+4y=94.

3. Convert x+y=35 to x=35-y, and substitute it into 2x+4y=94 to get y=12.

4. Substituting y=23 into x+y=35 yields x=23.

According to the process, there are 23 chickens and 12 rabbits in the cage. Blind search.

Considerations for solving equations.

1. If there is a denominator, go to the denominator first.

2. If there are parentheses, remove the brackets.

3. If you need to move the item, you will move the item.

4. Merge similar items.

5. The coefficient is reduced to 1 to obtain the value of the unknown.

6. Write "solution" at the beginning.

Suppose the number of chickens is x, and the number of rabbits is y

Because x+y=head.

2x+4y=feet.

So (2x+4y)-2(x+y)=foot-2 heads.

2y = foot spike return - 2 heads.

y=(foot-2 head) 2

The problem can be solved by the list method, the (hypothetical method) and the column equation method

So the answer is: hypothetical

Set up Xiao Mingxin's manuscript to answer x questions correctly and make wrong (10-x) questions.

Answer: 2x-(10-x)=14

Solution 2x-10+x=14

2x+x=14+10

3x=24x=8

The round shouts 10-x) = 2

Answer: Xiao Ming answered 8 slippery filial piety questions correctly and answered 2 questions incorrectly.

This question is one of the famous interesting questions in ancient China. About 1,500 years ago, this interesting question was recorded in the "Sun Tzu's Sutra". Here's how it is narrated in the book:

Today there are pheasants and rabbits in the same cage, there are thirty-five heads on the top, and there are ninety-four feet under it. The meaning of these four sentences is: There are several chickens and rabbits in the same cage, and counting from above, there are 35 heads; Counting from below, there are 94 feet.

How many chickens and rabbits are in the cage?

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

4x+2×(35-x)=94

4x+70-2x=94

2x=24x=12 35-x=35-12=23

A: There are 12 rabbits and 23 chickens.

1. Spiders have 8 legs, dragonflies have 6 legs and 2 pairs of wings, and cicadas have 6 legs and 1 pair of wings. There are 21 known three species of insects, with a total of 140 legs and 24 pairs of wings. How many of these three bugs are there?

Solution: If there are x spiders, then dragonflies and cicadas (21-x) are allowed. Because both dragonflies and cicadas have 6 legs.

8x+6×(21-x)=140

2x=14x=7

Dragonflies and cicadas: 21-7 = 14 individuals.

Solution: If there are x dragonflies, then there are only 14-x cicadas.

2x+1×(14-x)=24

x=24-14

x=10 so cicadas are: 14-10=4.

A: There are 7 spiders, 10 dragonflies, and 4 cicadas.

2. Spiders have 8 legs, dragonflies have 6 legs and 2 pairs of wings, cicadas have 6 feet and 1 pair of wings, and now there are 20 of these three insects, 126 legs and 24 pairs of wings. How many of each insect are there?

Solution: If there are x spiders, then dragonflies and cicadas (20-x). Because both dragonflies and cicadas have 6 legs.

8x+6×(20-x)=126

2x=126-120

x = 3 dragonflies and cicadas: 20-3 = 17.

Solution: If there are x dragonflies, then there are only 17-x cicadas.

2x+1×(17-x)=24

x=24-17

x=7 so cicadas are: 17-7=10.

A: There are 3 spiders, 7 dragonflies, and 10 cicadas.

Many chickens and rabbits are in one cage, and the rabbit has twice as many feet as the chicken, so if the chickens are x, then the rabbit has (number of heads - x), then the equation can be as .

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

An efficient solution to the chicken-rabbit cage problem.

Chickens and rabbits in the same cage formula.

Solution 1: (the number of rabbit's feet, the total number of feet, the total number of feet) (the number of rabbit's feet, the number of chicken's feet).

The number of chickens. Total number of chickens = number of rabbits.

Solution 2: ( Total number of feet Number of chicken's feet Total number of birds) (Number of rabbit feet Number of chicken's feet).

The number of rabbits. Total number of rabbits = number of chickens.

Solution 3: Total number of feet 2 - total number of heads = number of rabbits.

Total number of rabbits = number of chickens.

Solution 4: The number of rabbits = the total number of feet 2 - the total number of heads.

Total number of rabbits = number of chickens.

Solution 5 (Equation): x=( Total number of feet Number of chicken's feet Total number of chickens) (Number of rabbit feet Number of chicken's feet) (x=Number of rabbits).

Total number of rabbits = number of chickens.

Solution 6 (equation): x = :( number of rabbit's feet total number of feet total number of feet) (number of rabbit feet number of chicken's feet) (x = number of chickens).

Total number of chickens = number of rabbits.

Solution 7 The number of chickens = (4 The total number of chickens and rabbits - the total number of chickens and rabbits) 2 The number of rabbits = the total number of chickens and rabbits - the number of chickens.

Solution 8 Total number of rabbits = (total number of chickens and rabbits - 2 total number of chickens and rabbits) 2 Number of chickens = total number of chickens and rabbits - total number of rabbits.

Solution 9 Total number of legs 2 - total number of heads = number of rabbits Total number of birds - number of rabbits = number of chickens.

2 yuan and 5 yuan each x sheets, then 10 yuan has 50-2x sheets for a total of 240 yuan, so there is.

2x+5x+10(50-2x)=240

x=2050-2x=10

There are 20 pieces of 2 yuan.

5 yuan has 20

There are 10 pieces of 10 yuan for 10 yuan.

If you think it's good, give it satisfaction.

Solution: Let 2 yuan and 5 yuan each be x sheets, then 10 yuan is 50-2x sheets.

2x+5x+10 multiplied by (50-2x)=240 to get x=20

For 10 yuan, 50-2 times 20 = 10 sheets.

Answer: 20 tickets for 2 yuan.

20 pieces for 5 yuan.

10 for 10 yuan.

o( o haha

No, let's talk about it.