Two math problems, the first year of junior high school. It s fast!!

1. Solution: After B works for x days, A continues to process for y days, just to complete the task as scheduled. Then:

1/10）x＋（1/20）y＝1 ①

x＋y＝12 ②

Solve the equation to get x 8, y 4

That is, after B worked for 8 days, A continued to process for 4 days, just to complete the task as scheduled.

2. Solution: Let the time of the power outage be x hours, then:

1－x/4）＝2（1－x/3）

Solve the equation as: x 4 5 h.

This question can be done as a speed problem: the distance is the same, it is regarded as 1, and coarse: it takes 4h to complete the whole process, and the speed is 1 4;Fine: It takes 3h to walk the whole distance, and the speed is 1 3, so it is easy to understand.

1. Here you can directly set B to work for x days first, and A to continue processing for y days, just to complete the task as scheduled. Then:

1/10）x＋（1/20）y＝1

x＋y＝12

Solution. x＝8，y＝4

After B worked for 8 days, A continued to process for 4 days, just as the task was completed as scheduled.

2. Set the time of the power outage to be x hours, and there is:

1－x/4）＝2（1－x/3）

Solution: x hours.

Solution: Set the time of the power outage to x hours

4－x）／4＝2（3－x）／3

3（4－x）＝8（3－x）

12－3x＝24－8x

3x＋8x＝24－12

5x＝12x＝2．4

A: The power outage lasts for 2 to 4 hours.

Solution: If B works for x days first, then A needs to do (12-x) days.

1/10x+1/20（12-x）=1

x=8 then 12-x=4

Answer: If B works for 8 days first, then A needs to work for 4 days.

Solution: Suppose the duration of the power outage is x hours.

1－x/4）＝2（1－x/3）

x = A: The time of the power outage is hours.

1. After X days, A can continue to process before it can be completed. x 20 + (1 20 + 1 10) (12-x) = 1.

2, Let the power outage time of x hours, 1-1 4x=2 (1-1 3x), according to the ratio of the remainder is 2:1, and find x=hours.

Question 1: Let B do x days after A does (1 divided by 10) multiplied by x and then added (12-x) multiplied by (1 divided by 20) equals 1, and then solve it out, now there is no time to come back to you on the second day.

1.Suppose: The older one of the cypress trees is x years old, then:

x-(4000-x)=1000

2.The first one has no solution, x=2

The second is three-two.

1.The ages of the two trees are.

x+y=4000

x-y=1000

Add the two formulas to give 2x=5000

x=2500

y=1500

2.（1）2x=4 x=2

2）(x-2)(x-3)=0

x=2 or x=3

1、{ x+y=4000

x-y=1000

2. (1) x = 2 (not in parentheses).

2) Three-two.

1. Let the A tree be older, A tree + B tree = 4000

A tree - B tree = 1000

2-1.Subtract x from 1 on both sides at the same time to get 2x=4, so x=2 answers 2-2 in parenthesesAdding 4x to both sides gives us 6=4x, so x=3 2 (three-thirds).

1.There are two ancient cypress trees, one scholar said; The sum of the ages of these two ancient cypress trees is 4,000 years old; The age difference is 1000 years, if his statement is correct, please do the math, how old are these two cypress trees now?

Solution: Let one be x years old and the other be y years old, then there is an equation:

x+y=4000...1)x-y=1000...2)(1)+(2) gives 2x=5000, so x=2500;y=1500.

2.The following equations are given in parentheses to find the solution of the equation (1)3x+1=x,).

Solution: 2x=4, x=2

2）x-5x
）

Solution: 4x=6, so x=3 2

Let one be x and one be y, then there is x+y=4000x-y=1000, and the solution is x=2500y=1500, so the ages of the two trees are 2500 and 1500, 3x+1=x+5, 2x=4, x=2, x-5x+6= 0, x=2, 3

Question 1: Let one be x and one be y

So x+y=4000

x-y=1000

Question 2: Question 1: 2 (It doesn't seem to be, did you make a mistake).

Question 2:3

1.Let the age of the young ancient cypress tree be x

4000-2x=1000

2.One is all wood.

Multiply 3 times 2 times 4 minus 1

Multiply 1/2 + 2 times 1/3 + 3 times 1/4 + .n(n+1).

1-1/2+1/2-1/3+……1/（n-1）-1/n1-1/n

1。No, to tell the truth, you haven't learned square counting.

Multiply 1/2 + 2 times 1/3 + 3 times 1/4 + .n(n+1).

1-1/2+1/2-1/3+……1/（n-1）-1/n1-1/n

Multiply 1/2 + 2 times 1/3 + 3 times 1/4 + .n(n+1).

1-1/2+1/2-1/3+……1/（n-1）-1/n

1-1/n

y+4=2x-2

3x-1=1-2y

Solution: x=2

y=-2 If you don't understand something, ask again! Hope it works for you.

Establish a system of equations 1, y+4=2x, x-1=1-2y, and solve the system of equations. Do you know how to solve systems of equations?

Let the unit price of "Interpretation" be x, then the unit price of "Standard Dan Silver" is 25-x.

So the number of "Standards" bought by the school is 378 (25-x).

The number of early and late drafts of the "Interpretation" bought by the school is 1053 x.

So the number of math teachers = the number of Standards = the number of Interpretations.

That's 378 (25-x) = 1053 x.

The standard unit price is X yuan, and the number of buyers is y.

The unit price of the dry interpretation is x+25 yuan.

It can be obtained according to the title of the feast.

xy=378

x+25）y=1053

The solution is x=14

y=27, so the standard unit price is 14 yuan.

The unit price of the interpretation of Wang Zidong is 39 yuan.

The epitaph of Diophantus is a poem of harmony: 1 6 of his life was a happy childhood, and he lived another 1 12, and the thin beard climbed up his face, and after 1 7 of his life, he had a companion called Zi Hu, and he teased him for another 5 years, and he had a little angel, but the child's life was only half of his father's, and the old man who was overly sad suffered for 4 years, and he also said goodbye to the world.

In fact, Diophantus stated the length of his life in his epitaph:

The solution is: Set the loss of the figure to live to be x years old.

x=1/6x+1/12x+1/7x+5+1/2x+4x=25/28x+9

3/28x=9

x=84 So he lived to be 84 years old.

Age of marriage: 1 6 * 84 + 1 12 * 84 + 1 7 * 84 = 33

1.4/5 x-6 = -2

2.If two cars collide with each other at the same time, how long will it take for the two cars to meet each other? 45x+35x=40