A few simple elementary school problems, two very simple elementary school math problems.

1.Hurry up and speed up—— Hurry up and whip up the great achievement—— The terrain is flat—— One horse is flat and the river is huge—— Ten thousand horses gallop to expand the strength—— Recruit and buy horses There are many people and horses—— Thousands of troops and horses are uncertain - the heart ape is in the forefront - One horse takes the lead 2Solution:

Let the original price of commodity B be $x, then the original price of commodity A is (7 3) x [7/3)x+70]:(x+70)=7:

4, 7x(x+70)=[(7 3)x+70]x4, 7x+490=(28 3)x+280,(28 3)x-(21 3)x=490-280,(7 3)x=210,x=90,** of commodity A is (7 3)x90=210 Answer: ** of commodity A is 210 yuan, and ** of commodity B is 90 yuan.

Hurry up and speed up—— Non-stop and make great contributions—— Sweat and horses, the terrain is flat- One horse is flat, the momentum is huge—— Ten thousand horses gallop to expand their strength- Recruit and buy horses, there are many people and horses- Thousands of troops and horses are uncertain - the heart ape is in the forefront - Single-handedly use the difference to solve.

Originally, A accounted for 7 (7-3) of the ** difference, and now A accounts for 7 (7-4) of the ** difference, and the ** difference is regarded as"1", first seek ** poor, and then seek A.

A: 70 [7 (7-4)-7 (7-3)] 7 (7-3)=210 yuan.

B: 210 7 3 = 90 yuan.

I only know 1Hurry up and speed up- Non-stop and make great contributions- Sweat and horses, the terrain is flat, one horse is flat, the momentum is huge, ten thousand horses gallop to expand strength, recruit troops, buy horses, there are many people, thousands of troops and horses are uncertain, and the heart ape is in the forefront - single-handedly 2Let a be $x.

x + 70 times (70 + 3 7x) = 7 4

x=210a210 yuan.

b = 210 times 3 7 = 90 yuan.

Fast horses, sweat and horse merit, a horse Pingchuan, thousands of horses galloping, (busy), recruiting troops, thousands of horses, heart ape horses, a horse in the lead...

Let a=7x, then b=3x

7x+70）/（3x+70）=7/4

x=30 yuan.

A commodity: 30 * 7 = 210 yuan.

B commodity: 30 * 3 = 90 yuan.

Hurry up and work hard.

One horse and ten thousand horses galloping.

Recruiting Horses is like a dragon.

The heart of the ape. Lead the way.

Let a be $x.

x+70/(70+3/7x)=7/4

x=210a210 yuan.

b = 210 * 3 7 = 90 yuan.

1.Hurry up and work hard.

One horse and ten thousand horses galloping.

Recruit thousands of troops.

The heart ape is the first horse.

2. Let a be $x.

x+70/(70+3/7x)=7/4

x=210a210 yuan.

b = 210 * 3 7 = 90 yuan.

Thousands of troops, horses

The first inscription.

Column equation system: bai

x + y + z =860

x/3 = y/4

y/5 = z/2

Answer: dux = 300, y = 400, z = 160 Second question zhi: (x - x - 190) ( x + 190) = 3 5, answer.

DAO case: x=400

1. Plant x willow trees, Y poplar trees, and Z locust trees.

x+y+z=860

1/3)x=（1/4)y

1/5)y=（1/2)z

x= z=160

2.Let the total number of loaded trees be x

15%x+190+(3/5)x=x

x=760

1. Elementary school.

It seems that you can't solve the system of three-self one-dimensional equations, and analyze it with ratios and proportions:

1/3 of the willow du is equal to 1/4 of the poplar zhidao, i.e., willow:poplar 3:4 i.e.:

Willow Tree 3 4 Poplar Tree 1/5 of a poplar is as much as 1/2 of a locust tree. i.e., poplar: locust tree 5:

2 i.e.: locust tree tree 2 5 poplar tree.

Next, you can solve the equation in a column:

There are x poplars, 3 4 x willows, 2 5 x locust trees (1 + 3 4 + 2 5) x 860, x 400 ......Poplar.

3/4x＝300……Willow.

2/5x＝160……Tree.

2. Analysis: because, unplanted: planted 3:5 i.e., the number of unplanted trees 3 5 The number of planted trees is set as: the total number of trees is x, then the first day of the tree, the unplanted 3 5 (tree column equation: (1+3 5) (

Solution: x 400 (tree).

We are happy to answer for you.

First, the elementary school seems to be unable to solve the ternary system of equations, and use the dao ratio and proportion to analyze: back.

1/3 of a willow is equal to 1/4 of a poplar.

Answer, i.e., willow: poplar 3:4 i.e., willow tree 3 4 poplar tree 1/5 of a poplar tree is as much as 1/2 of a locust tree. i.e., poplar: locust tree 5:2 i.e., locust tree tree 2 5 poplar tree tree.

Next, you can solve the equation in a column:

There are x poplars, 3 4 x willows, 2 5 x locust trees (1 + 3 4 + 2 5) x 860, x 400 ......Poplar.

3/4x＝300……Willow.

2/5x＝160……Tree.

2. Analysis: because, unplanted: planted 3:5 i.e., the number of unplanted trees 3 5 The number of planted trees is set as: the total number of trees is x, then the first day of the tree, the unplanted 3 5 (tree column equation: (1+3 5) (

Solution: x 400 (tree).

1. There are x willow trees, Y poplar trees, and z locust trees. Then there are:

x+y+z=860; 1\3x=1\4y;1 5y=1 2z gives y=400, x=300, z=160

2.There are x trees in total.

x-15%x-190) (15%x+190)=3 5 solution gives x=400

The long jump results of the three students were: A 4 and 3/8 meters, B 4 and 9/9 4 meters, and C 4 and 1/3 of a meter.

B 4 and 4/9 of a metre "A 4 and 3/8 of a metre" C 4 and 1/3 of a metre .