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There is a distance of 160km between a and B. three hours after a long-distance bus leaves from a, a car also leaves from A. as a result, the car arrives at B 20 minutes later than the long-distance bus. It is known that the speed of the car is three times that of the long-distance bus. The speed of the two cars is calculated
Suppose the speed of the coach is XKM / h, then the speed of the car is 3xkm / h. according to the meaning of the question, we get 160x = 1603x + 3 − 13, and the solution is x = 40. After testing, x = 40 is the root of the original equation, and the root of the original equation is x = 40. The speed of the car is 40 × 3 = 120km / h. answer: if the speed of the coach is 40km / h, then the speed of the car is 120km / h
Application of fraction equation in grade one of junior high school
If Party A does a certain job alone, it will be completed on time; if Party B does it alone, it will exceed the specified date by three days. Now, after three days of cooperation between Party A and Party B, the rest of the work will be completed by Party B alone, and it will be completed on time. So, how many days is the specified date for completing this work?
Party A and Party B cooperate for two days.
Set to X days
Then a's progress is 1 / X
B is 1 / (3 + x)
that
1-[1/X+1/(3+X)]*2=1/(3+X)*(X-2)
The solution is x = 6
The expansion of the application of fractional equation in grade one of junior high school
For a project, the number of days required by team a is a times of that required by team B and team C, the number of days required by team B is B times of that required by team a and team C, and the number of days required by team C is c times of that required by team a and team C
Find the value of - + - +
a+1 b+1 c+1
X=(Y+Z)*a
a=(Y+Z)/X
1/(1+a)=1/[1+(Y+Z)/X]=X/(X+Y+Z)
Similarly: 1 / (1 + b) = Y / (x + y + Z)
.
1/（1+a）+1/（1+b）+1/（1+c）=1
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