Suppose A has done X days and B has done Y days, and the total project is 1 (analysis: this 1 is not a certain total project number, but a virtual quantity under uncertainty, representing a whole).

Then two formulas are given: x+y = 18, (1/20) x+(115) y =1.

(Analysis: 1/20 represents the speed that A does every day, and115 represents the speed that B does every day. Multiply it by the number of days when two people do X and Y respectively, and then add it up. At first, the total number of assumed projects is 1).

Through the above two formulas

X+Y= 18，( 1/20)X+( 1/ 15)Y = 1

Calculate X= 12 Y=6.

A: A did it for 12 days and B did it for 6 days.

2. solution: suppose there are x minutes before school.

(X+8)*50=(X-5)*60

X=70

(analysis: the distance from home to school is fixed, so it is upright. The left side moves at a speed of 50 meters per minute. Because it's eight minutes late, it takes eight more minutes to walk this way. X+8 is time, 50 is speed, and multiplication is total distance. The right side is moving at a speed of 60 meters per minute. Because it is five minutes early, the five minutes saved should be cut off. X-5 is time, 60 is speed, and multiplication is total distance. The total distance between the two sides is equal, so connect them with an equal sign to establish and solve this one-dimensional linear equation. )