"Practical Problems and Systems of Linear Equations of Two Variables" System of Linear Equations of Two Variables PPT Courseware 2

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"Practical Problems and Systems of Linear Equations of Two Variables" System of Linear Equations of Two Variables PPT Courseware 2

Wukong followed the wind to explore the traces of the demon, and it only took four minutes to travel a thousand miles.

When we return, we will reach 600 in four branches. How fast will the wind be?

Downwind speed = Wukong walking speed + wind speed

Headwind speed = Wukong walking speed - wind speed

Solution: Suppose Wukong’s walking speed is x miles per minute and the wind speed is y miles per minute, then

According to the topic

4(x+y)=1000

4(x-y)=600

x=200 y=50

Answer: The wind speed is 50 miles per minute.

strive to improve yourself

Exercise 1: An 18-meter-long steel material needs to be sawed into 10 sections, and the length of each section can only be one of two models: "1 meter or 2 meters". Xiao Ming estimates that there are 3 sections for 2 meters. Do you think he is right? precise? why? How many sections should be taken for 2 meters and 1 meter?

Exercise 2: There are 5 large restaurants and 2 small restaurants in a certain university. After testing: 1 large restaurant and 2 small restaurants are opened at the same time, which can accommodate 1680 students; 2 large restaurants and 1 small restaurant are opened at the same time. It can accommodate 2,280 students.

(1) How many students can dine in a large restaurant and a small restaurant? (2) If 7 restaurants are open at the same time, please estimate whether they can serve the entire school’s 5,300 students. Please explain why.

A system of equations represents the quantitative relationships in each of the following problems:

(1). The percentage of iron contained in type A ore is 1.5 times that of type B ore. The ore mixed with 5 parts of type A ore and 3 parts of type B ore contains 52.5% iron. Suppose type A is x% and type B is y%, then x%=1.5y% 5·x%+3·y%= (5+3) · 52.5%

(2) Two pieces of aluminum-tin alloy, the first piece contains 40 grams of aluminum. Contains 10 grams of tin, and the second piece contains 3 grams of aluminum. 27 grams of tin, to obtain 40 grams of alloy containing 62.5% aluminum, take the first piece as x grams and the second piece as y grams,

Then x+y=40 40/40+10·x+3/3+37·y=62.5%×40

(3)A. Take 100 grams of each type of brine B and mix them. The salt content of the resulting brine is 10%. If 400 grams of type A salt water and 500 grams of type B salt water are mixed, the resulting salt content will be 9%. Let A be x% and B be y%,

Then 100 ·x%＋100 ·y%＝2×100×10% 400 ·x%＋500 ·y%＝(400＋500) ·9%

Ask students to review the concepts they learned in this lesson about concentration issues

The basis is: solution = solute + solvent solute = concentration × solution

The equivalent level is:

The sum of the solutions before mixing = the solution after mixing

Sum of solutes before mixing = solute after mixing

Solving word problems involving a system of equations must also be tested, not only by substituting into the system of equations, but also by substituting into the problem for testing.

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For more information about the PPT courseware "Practical Problems of Systems of Linear Equations of Two Variables and Systems of Linear Equations of Two Variables", please click on the practical problems of systems of linear equations of two variables ppt and Systems of Linear Equations of Two Variables ppt tag.

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Update Time: 2024-07-11

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