"Application of linear equations of one variable" PPT courseware 3

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"Application of linear equations of one variable" PPT courseware 3

What is the general process for solving practical problems using equations?

1. Review the question: analyze the meaning of the question and find out the quantities and their relationships in the question; review

2. Set the element: Choose an appropriate unknown number represented by a letter (such as x) and set it

3. List equations: list equations based on equivalence relationships; list

4. Solve equations: find the value of the unknown; solve

5. Verification: Check whether the obtained value is correct and consistent with the actual situation, and write the answer; verification

6. Answer: Answer the requested answer. answer

I see

distance = speed x time

Speed ​​= distance ÷ time

Time = distance ÷ speed

encounter problem

Question 1

The distance between stations A and B is 450 kilometers. A local train leaves station A and travels 65 kilometers per hour. An express train leaves station B and travels 85 kilometers per hour.

Find (1) How many hours will it take for two cars to meet each other when they drive out at the same time and go towards each other?

(2) The express train drives first for 30 minutes, and the two cars are traveling in the opposite direction. How many hours does the slow train travel before the two cars meet?

(1) Solution: Suppose two cars leave at the same time, go towards each other, and meet each other in x hours.

65x+85x=450

Solving the equation gives x=3

Answer: Two cars drove out at the same time, heading towards each other, and met each other in 3 hours.

(2) Solution: Suppose the express train starts driving for 30 minutes, the two cars are moving towards each other, and the slow train meets each other after driving for x hours.

30 minutes = 0.5 hours

85×0.5+85x+65x=450

Solve for x=

Method 2: 85(0.5+x)+65x=450

Running water problem

The time it takes for a ship to sail from place A to place B along the current is 1.5 hours less than from place B to place A against the current. It is known that the speed of the ship in still water is 18 kilometers/hour and the speed of the current is 2 kilometers/hour. What is the distance between A and B?

Equivalent relationship: distance along the current = distance against the current

Water speed: the speed at which water flows by itself

Ship speed: the speed of a ship in still water

Downstream speed = still water speed + water speed

Speed ​​against water = speed in still water – speed of water

Solution: Suppose it takes x hours for a steamboat to sail upstream from point B to point A.

(18+2)(x-1.5)=(18-2)x

x=7.5

(18－2) ×7.5=120

Answer: The distance between A and B is 120 kilometers.

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Update Time: 2024-09-14

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