Binary Linear Equations and Practical Problems

A shopping mall buys a garment for each piece of x, if it is stipulated that it is sold at each piece of y, an average of 15 pieces are sold per day, and a total profit of 22,500 is made in 30 days, in order to raise funds as soon as possible, the mall decides to sell each piece at a price of 20%, and the result is an average of 10 pieces per day than before the price reduction, so that 30 days can still make a profit of 22,500 yuan, and the value of x, y.

y-x=22500/30/15

4y/5-x=22500/30/25

Solution x=50

y=100 After a shopping mall buys two kinds of clothing A and B, the price is increased by 40% **, during the "Spring Festival" shopping mall to engage in discounts**, it is decided to pay a hundred percent off the price of two kinds of clothing **, a customer buys A and B Liang clothing to pay a total of 182 yuan, the sum of the two clothing prices is 210 yuan, ask what the purchase price and the price of these two kinds of clothing are.

Let the price of A be x and the price of B be y

x+y=210

Solution x=70

y=140, the purchase price A is 50

B is 100

y-x=22500/30/15

4y/5-x=22500/30/25

Solution x=50

y=100 Let the price of A be x and the price of B be y

x+y=210

Solution x=70

y=140, the purchase price A is 50

B is 100

Solution: Set up and buy x first-class seats, y second-class seats, and z third-class seats.

First class and second class First class and third class {300x+200y=5025 x+z=36

x+y=36② 300x+125z=5025

The solution of this square multiplier is x= The solution of this square multiplier is x=3 z=33

This option does not work, this plan can be 2nd class and 3rd class.

y+z=36

200y+125z=5025

The solution of this square is z=29 y=7

This scenario does.

A: There are two options for the garment company to choose from, buy 3 first-class seats, 33 third-class seats, buy 7 second-class seats, and 29 third-class seats.

If the number of first-class, second-class, and third-class seats is x, y, and z respectively, then 300x+200y=5020 and x+y=36 are combined and taken as an integer, and the solution x<0 is not possible. In the same way, the variables are x, z, y, and z to solve x=3 and z=33 respectively. y=7 and z=29.

There are three ways: 1. Choose a first-class seat and a second-class seat: 200 (36-x) + 300x=5020, get x=, it is impossible to decimally.

2. Choose the third class and the second class: 200 (36-x) + 125x = 5020, it is impossible to get a decimal number.

3. Choose the first class and the third class: 125 (36-x) + 300x = 5020, x is still a decimal, so this is impossible.

Grand prize, first prize for 36 customers ** finals,

Solution, set the first-level tomatoes have x catties, and the second-level tomatoes have y catties, get:

8x+6y==720

x+y)*7==720-20

Solve the system of equations and get:

x==60 (catty).

y==40 (catty) so get:

In this way, the ** of tomatoes is cheaper than the ** of the first one, so the total sales of the second type of tomatoes are 20 yuan less than the first one.

In fact, there is no need to calculate, less 20 yuan, that is, the first-class tomatoes are 20 catties more than the second-grade tomatoes, 20 catties are 160 yuan, 720-160 = 560, 560 7 = 80;That is, 40 each for the first level and 40 for the second level. Finally, level 60 is known; Level 2 40

Solve x rooms, y people.

According to the meaning of the question, y-7x=7

y=9(x-1)

then x=8, y=63

There are 8 rooms and 63 guests.

The correct answer is: B

The solution is as follows: Let the velocity of the ship in still water and the velocity of the current be x,y kilometers and hours respectively, and list the following equations according to the meaning of the problem:

7（x+y）=140 （1）

10（x-y）=140 （2）

Solving this system of equations yields x=17, y=3

Let the single digit be x and the ten digit be y

10y+x-(x+y)=9 (1)

10x+y-10y-x=27 (2)

y=1 from (1).

Substitution (2).

The two-digit number x=4 is 14

The original plan was to be completed in x days, and this batch of farm tools was y.

20x+10=y (1)

23x-20=y (2)

1) substitution (2) obtained.

23x-20=20x+10

x=10,y=210

210 pieces of farm tools, originally planned to be completed in 10 days.

Let the number in the 10th place be m and the number in each place be n;

10m+n-(m+n)=9m=9, so m=110n+m-(10m+n)=9*(n-m)=27;

So n-m=3;So n=4

Class 1 and Class 5 scored X and Y points respectively.

then x:y=6:5

x=2y-40②

Substituting into , we get (2y-40): y=6:5

Then 5 (2y-40) = 6y

10y-200=6y

4y=200

The solution is y=50

Substituting , we get x=100-40=60

So Class 1 got 60 points, and Class 5 got 50 points.

Let the score of one class be x and the score of class five be y

x：y=6：5

x=6y/5①

x+40=2y②

Substitute 6y 5+40=2y

4y/5=40

y=50 and x=60

Let class 1 score x and class 5 score y

x/y=6/5

x=2y-40

Solution: If class 5 gets x points, then class 1 gets y points.

y/x=6/5

y-x=2x-40