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Mathematics in Grade One and Practical Problems in One-variable Linear Equation
1. Solution: If the purchase price of Party A is X yuan, the purchase price of Party B is (600-x) yuan.

A linear equation [x (1+50%)+(600-x) (1+40%)] × 90%-600 =174 is obtained from the problem.

Solution: x=200

600 x = 400

2.( 1) can be obtained from the question: y=400-(x-40)×80.

y=400-80x+3200

y=-80x+3600

(2) Substitute 380 barrels into (1) to get the relation: 380=-80x+3600.

-80x=-3220

x=40.25

Then: the total annual cost of this kind of purified water is: 40.25×380+780= 16075 yuan.

The total cost of drinks is: 50× 120=6000 yuan.

So it is more cost-effective to use drinks.

3.( 1) Solution: Let the painting area of each room be x square meters.

Then: [(4x-44)/3]-10 = (6x+14)/7.

Solution: (4x-74)/3=(6x+ 14)/7.

28x-5 18= 18x+42

10x=560

x=56

(2) According to the problem, the daily painting area of the first-class mechanic is: (56× 4-44) ÷ 3 = 60m2.

Secondary mechanic: 60- 10 = 50m2.

For every Y first-class technicians, there are 20-y second-class technicians.

Then: 60y+(20-y)50=20×56.

60y+ 1000-50y = 1 120

10y= 120

y= 12

20-y=8

Question 4: I really lack the conditions about traffic flow, or I can't do it.