Primary 6 Lessons 10

You should deeply understand the meaning of the topic, and you should not draw too many pictures.

A little bit of experience in solving the application problem:

2. Clever unknowns. Several quantities can be set as unknown in a worksheet, but which one is easier should be carefully considered. For example:

The ratio of the velocity of A and B is 3:2, and when finding the velocity of A and B, we can let the speed of A be a kilometer hour and B be b kilometers and hour, which is a system of binary linear equations; Or let the velocity of A be a kilometer-hour, then B is 2 3a kilometer-hour, so that although it is a one-dimensional equation, there are fractions; Or let A have a speed of 3a kilometers per hour and B a speed of 2a kilometers per hour.

It can be seen that the last one tries to be the best. Unknowns are set according to different topics.

3. List equations according to the equiquantity relationship.

4. Solve equations. At this time, we may encounter two unknowns, and we can only list one equation, we need to see if there are still implicit conditions, such as the number of people and the number of objects, all of which must be positive integers, which are implicit conditions, especially in inequality equations. There is also the need to test the root of fractional equations.

5. Write down the unit and answer. This step is often overlooked, but in fact, it is a reflection of whether you have read the question, whether you know what the question requires, and whether you have to stand for marks in the exam.

6. Practice diligently, practice makes perfect. Touching the analogy bypass, drawing inferences from one another.

This is a little bit of my personal experience in docking application problems, I hope it will be helpful to you. A little bit of experience.

The topic should be carefully examined and the mind should be clear.

The first side of the silver first found Cong Sakura 42 and 36.

Greatest common factor.

6. The side length of the small square is 6 cm. Number of sheets.

42 sheets.

4 (4+3) = 4 out of 7.

30 (4/7 - 2/2 liter 1) = 420 (km).

I have also been at the mountain level for 6 years, and this noisy question is not difficult.

According to the topic, two conditions are derived:

1.4/5A + 3 = B + 1/5A.

2.One-third B + A-7 = two-thirds B.

Integrate the conditions:

3.Three-fifths A + 3 = B.

4.A-7 = one-third B.

Then change the 4 conditions to get.

5.3A-21=B.

A(21+3) (3-3 5)=10 (pieces) B9

B Yes. (7-2+3) (4 3-2 3) = 9 Suppose B does not have these 3 and is equal to B. That is to say, B has 6 A, and when A gives 1 to himself, 5, B is equal to A.

So B is 2 5 less than A. That is, 3 5 of A. A is:

6 (1-1 5 2) = 10 pieces. 10 for A, 9 for B.

Is there any other solution than the system of equations?

It can be inferred from a system of equations, that is, two systems of equations are transformed to reduce one unknown number to become a formula with only one unknown. In fact, I didn't learn it, and I didn't say that I wouldn't use it, and it wasn't a high-end knowledge.

Draw a line segment diagram. When you draw frame A and frame B separately.

The difference is divided by the score paired.

And he begged for it.

The resulting solid line is drawn.

Take someone else's dotted line.

It's easier to solve with equations.

If it really doesn't work, I'll give you the answer directly:

A:10 B:9

There are x pieces of sugar in the armor basket.

1\3（x-1\5x+3-1\5x)+x-7=(1-1\3)(x-1\5x+3-1\5x)

Solution: x=7

This is not a system of equations.

I found that the current elementary school topics are really difficult.