How to use a one dimensional equation in a column

This is relatively simple, for example: x=8, y=9, these are all unary equations.

Read the abstract of the example problem carefully, clarify the meaning of the question, find out an equal relationship that can represent the full meaning of the application problem, the process of analysis can let the students only write on the draft, in the process of writing the solution, ask the students to set the unknowns first, and then list the required algebraic formula according to the equality relationship, and then express the equality relationship into the form of an equation, and then solve the equation, and write the answer, when setting the unknown, if there is a unit, you must ask the students to write it after the letter, as in example 1, you can't write "let the original have x kg of flour" as " Let there be x ". In addition, in a column equation, the units of each algebraic equation should be the same, as in Example 1, the units of the algebraic equations "x""—15%x" and "42500" are all kilograms. In this example, the key is to find this equality relation, make one of the numbers involved in it an unknown number, and list the rest of the numbers in an algebraic equation with a known number or an algebraic equation containing both known and unknown numbers.

The equality relationship in Example 1 is simple and obvious, and students can find it out by themselves through heuristics.

Read the question carefully, set the quantity as an unknown number, find out the equation of the relationship according to the common sense of life, and finally solve the equation.

It is best to be able to find two equal relations, one with unknowns and one for column equations, note: one equal relation cannot be used twice!!

Let an x, and an equal equivalence relation!

Generally, the problem is set to unknowns, find the relationship between equal quantities, and list equations.

The first thing to do is to find the equiquantity relationship

The general steps for solving a problem with a one-dimensional equation.

1) Examination of the topic: clarify the meaning of the topic

2) Find the equivalence relation: find the equivalence relation that can represent the meaning of the problem (3) Set the unknowns and list the equations: After setting the unknowns, express the relevant equations with letters, and then use the found equivalence relations to list the equations

4) Solve the equation: solve the listed equation and find the value of the unknown (5) Test and write the answer: Check whether the value of the unknown is the solution of the equation and whether it is in line with reality, and write the answer after the test.

First: Review the topic Second: Find the equivalent relationship Third:

Find the unknowns and list the equations Fourth: solve, test and answer. Note:

The solution depends on the nature of the equation. 1) Examination of the question: It is necessary to clarify what is known, what is not known and its interrelationship, and use x to represent a reasonable unknown in the question.

2) Find an equal relationship that can represent the full meaning of the problem according to the meaning of the question. (3) According to the equality relation, the equation is listed correctly, that is, the equation listed should satisfy the equal quantity on both sides of the equal sign; The units of the algebraic equation on both sides of the equation should be the same.

4) Solve the equation: find the value of the unknown.

5) Write out the answer clearly and completely after the test. The test should be that the solution obtained by the test will not only make the equation true, but also make the application problem meaningful.

Question: Can you talk about how to make a one-dimensional equation and make it simpler, and the child will not be able to list a one-dimensional equation.

Quantitative relations in practical problems: general and difference relations or multiples relations, often used "altogether", "ratio......more", "than ......Less", "Yes......When solving the problem, you can grasp these terms to find the equivalent relationship, list the equations in the order of description, ask questions mainly about distance and sales, and now the answer is to find the equivalent relationship according to the common quantity relationship.

Common quantity relationships: work efficiency, working hours, total work; Unit price, quantity, total price; The speed of the time of the distance ......For example: "The retail price of a certain style of clothing is 36 yuan for 1 set, and the existing price is 216 yuan, how many sets of clothes can I buy in total?"

According to the quantity relationship of "unit price, quantity, total price", equation 36 216 can be listed

Asking questions like this is too easy for you.

And the process is clear.

1. Solution: Set the car to be x meters away from the valley at this time. 72 4 3600 + 2x = 340 4x = solution:

Set up a store to buy x pieces of clothing. (300 30-2) (x-30)=380-300x=403, solution: with x block sugar.

x-1=(x+2) 2x=4 children: 4-1=3 (person) 380-6 40=120 (yuan)3

The number of "blisters" is 3 times that of "pearls".

The number of "Asahi Dragon" is twice that of "Pearl".

There are x pearls.

Then the number of "blisters" is 2x

The number of "Asahi Dragon" is 3x

Because "Chaotian Dragon" has 1 less than "Blister".

So 3x-2x=1

x=1so.

There is 1 pearl.

Then the number of "blisters" is 2

The number of "Asahi Dragon" is 3

If the pearl is x, then the blister is 3x, and the dragon is 2x

3x-2x=1

x=1 pearl=1

Blisters = 1 * 3 = 3

Chaotian Dragon = 1 * 2 = 2

Solution: The average amount of dust trapped in a year is x milligrams for a single leaf of a Chinese acacia tree.

The average amount of dust collected in a single ginkgo leaf is 2x-4 mg a year.

11（2x-4)=20x

22x-44=20x

22x-20x=44

2x=44x=22

That is, the average amount of dust trapped in a leaf of a Chinese acacia tree is 22 milligrams in a year.

Hope you do!

I'm sorry, I'll make a binary equation.

1.(1) Solution: If doe is x°, then 2x 60, x 30. Answer: DOE is 30°

2) Solution: If doe is x°, then 2x 180, x 90. Answer: DOE is 90°

3) Solution: If doe is x°, then 2x 270 or 2x 90, x 135 or 45. A: Set the doe to 135° or 45°.

2.(1) Solution: Set up a supermarket bookcase x yuan, then (210 20) x 20 80 (210x 70y), x:

y＝3:4。A:

When the ratio of the number of bookcases to bookshelves is not higher than 3:4, it is advantageous to buy it at A supermarket.

2) Solution: Because 20:100 1:

5 3:4, so it is advantageous to buy it at supermarket b. Therefore, 20 80 (210 20 70 100) 179200 (yuan).

A: At least 10,000 yuan should be prepared.

3.(1) Solution: It takes x y hours for a team member to catch up with the convoy, then 35x 55y 0, x:y 11:7, x y 11 7 4. A: It takes 4 hours for Team A to catch up with the convoy.

2) Solution: Set up team member B (20 45 x) hours later to be able to meet with the team. 20 45 (45 35) 20x 1, x 28 81, 20 45 28 81 64 81 (hours).

Answer: Team Member B 64 81 hours can catch up with the convoy.

Solution: Let the first speed x km/h, and the B velocity (x-2) km/h 5x+5(x-2)-(2x+2(x-2))=726x=78

x=13x-2=11

Distance from school to railway station = (13 + 11) * 2 + 36 = 84km

Set A x km h, B x h.

5x+5(x-2)-(2x+2(x-2))=726x=78

x=13x-2=11

Distance from school to railway station = (13 + 11) * 2 + 36 = 84km

Set the correspondent x minutes to catch up with the student team, by the topic, there is an equation:

5×(18+x)=14x

Solution x = 10 minutes.

The correspondent has 10 minutes to catch up with the student team.

18 minutes = hours, let x hours catch up with the team, according to the equation of the problem: 5 solve: x = 1 6 hours = 10 minutes.

Arithmetic solution: 5 km (team is km before the starting point).

14-5=9 km (a bicycle is 9 km faster than a person).

hours) = 10 (minutes).

5 kmh = (250 3) m min.

14 kmh = (700 3) m min.

1500 150 = 10 minutes.

Solution: Set up a correspondent X points to catch up with the student team.

Because the student team walked 18 minutes at a speed of 5 km, they walked 5 18 = 90 km equation: 5 x 90 14 x solution x 10

Answer: The correspondent can catch up with the student team in 10 points.

If the previous month's fee is 1, the current month's fee is (1+4%). If the oil ** of the previous month is y, then the import volume of the previous month is 1 y, the import volume of this month is 1*(1-5%) y, the ** of the current month is (1+4%)*y (1-5%), and the growth rate is [(1+4%)*y (1-5%)-y] y

Let A be x per day, then B is x+1.

A has more days than B.

So 30 x-30 (x+1)=3 2

Denominator 20(x+1)-20x=x(x+1)x +x-20=0

x+5)(x-4)=0

x>0, so x=4

30 x = A: A does 4 per day, and the original plan is to complete it in one day.

If A makes x parts per day, then B makes x + 1 per day;

A takes 30 x days to complete, B takes 30 (x+1) days to complete;

B is completed one day in advance, then:

30/x=30/(x+1)+

x=4 then worker A makes 4 parts a day and is originally planned to be completed in one day.

Let's say A does x per day, so B does x + 1 per day, and the number of days originally planned to be completed is: 30 x according to the question:

Equation: 30 x-30 (x+1)=

Solution: x1=-5 (rounded).

x2=4 Planned number of days to complete: 30 x=30 4=

Solution: The armor can make x parts per day, and the original plan was to be completed in y days.

x+1）·(