Solve 1 math problem. Around the first day of junior high school, the process is required.

To build a road, Team A completed it in 15 days and Team B completed it in 12 days. After the two teams worked together for 4 days, Team B was transferred, and Team A continued to repair the rest of the road. How many days does Team A have to repair? How many days did he practice?

Solution: Team A completes 1 15 of the project every day, Team B completes 1 12 of the project every day, and A and B jointly complete 1 15 + 1 12 = 3 20 of the project every day. After the two teams repaired for 4 days, the remaining amount of work is 1-4*3 20=2 5.

Team A needs to complete the remaining amount of work alone: 2 5 divided by 1 15 equals 6 days. So Team A still has 6 days to repair, and he has a total of 4+6=10 days.

Solution: Team A's efficiency: 1 15 = 1 15

Team B Efficiency: 1 12 = 1 12

The efficiency of the two teams: 1 15 +1 12 = 3 20 20 to complete in 4 days: 3 20 4 = 3 5

Amount of work remaining: 1-3 5 = 2 5

The remaining team A needs days: 2 5 1 15= 6 (days) Team A has repaired: 4+6=10 (days) Team A repairs 1 15 a day, and team B repairs 1 12 a day.

6 + 4 = 10 (days).

A: Team A spent a total of 10 days.

Solution: Set up team A to repair 1 15 a day, and team B to repair 1 12 a day.

6 + 4 = 10 (days).

A: Team A spent a total of 10 days.

Solution: The first team has been on for a total of x days, which can be obtained from the title:

4/12+x/15=1

solution, get: x=10

ps:It's pretty simple.。。

Solution: The first team has been repaired for a total of x days, according to the title:

4/12+x/15=1

x = 10 (days).

A: Team A spent a total of 10 days.

6 days, A working rate 4 5y

B Working rate y

12y-（4y/5+y）x4}/（4y/5）=6

Team A has been repaired for a total of x days, (1 15) x + 4 * (1 12) =

If the merchant arranges x and 10-x trucks for A and B, then 4x+2*(10-x) 30

2x+3*(10-x)≥23

Solve 5 x 7

Therefore, there are 3 kinds of schemes when the business arranges two kinds of trucks A and B: 1, each of the two kinds of trucks A and B; 2, A, B two kinds of trucks; 3. There are two kinds of trucks A and B.

Set up the merchant to arrange a type of truck x two.

4x+2(10-x)≥30

2x+3(10-x)≥23

5 x 7 and x is an integer.

There are three options.

Arrange a car 5, 6, 7 taels.

If car A x is used, car B will be 10-x. According to the inscription, the inequalities are grouped as.

4x+2(10-x)" 30, 2x+3(10-x)" 23, the solution is 5"x"7, so the scheme is to use 5 cars A and 5 cars B; 6 cars A and 4 cars B; 7 cars A and 3 cars B.

Let car A x and car B y, and it can be seen that the three equations of x+y=10, 4x+2y=30, 2x+3y=23 are satisfied, so there is only one set of solutions, that is, 5 cars for each of the two cars, which is hoped to be adopted.

Solution: If truck A rents x vehicles, then truck B rents (10-x) vehicles.

4x+2（10-x）＞30

2x+3（10-x）＞23

Answer: ......

Do the math yourself.

A time slot of 7:50-8:00 can be set according to the topic.

The number of motorcycles in the internal system.

is 5x, and the number of trucks in the period of 8:bai00-8:10 is 4x.

According to **, the number of motorcycle dao cars in the period of zhi8:00-8:10 is 30-5x, and the number of trains in the period of 7:50-8:00 is 20-4x.

Then the system of equations can be listed: 5x+7+20-4x+12=44, 30-5x+7+8= Bring x=5 in to find it.

a^2+2a-1=0,a+1)^2-2=0

a+1)^2=2

a+1=±√2

a=±√2-1

When a = 2-1, substitution.

a 2+1 a 2 ( 2-1) 2+1 ( 2-1) 2=2 When a = - 2-1, substitute.

a^2+1/a^2＝（-2-1）^2+1/（-√2-1)^2=4-2√2

a+1 a) 2=a 2+1 a 2+2=4 or 6-2 2

x10: 5 (2 x + 3y) = 2 (3 x + 2 y) + 20 x10 : 5 x3 (2 x + 3 y) = 2 x 2 (3 x + 2 y) + 60 is reduced to :

10x+15y=6x+4y+20②：30x+45y=12x+8y+60

Re-simplified. x10: 5 (2 x + 3y) = 2 (3 x + 2 y) + 20 x10 : 5 x3 (2 x + 3 y) = 2 x 2 (3 x + 2 y) + 60 is reduced to :

10x+15y=6x+4y+20②：30x+45y=12x+8y+60

Re-simplified. ：4x+11y=20

18x+37y=60

x9- x2; （36x+99y）-（36x+74y）=180-120

25y=60

y = 12/5

Substitute y=12/5 into 4x+11x12/5=20x=minus 8/5

x = minus 8/5 y = 12/5

Punch it out without missing a step, enough detailed drops! Hit me for half a day ==.

Let (2x+3y) 2=m and (3x+2y) 5=n.

Change the system of equations to m=n+2

3m=2n+6

The solution is m=2 n=0

Bring in the set-up. 2x+3y=4

3x+2y=0

The solution is x=-8 5 y=12 5

x3- Got:

0=3x+2y

Substitute back: 2x+3y=4

Now it's easy.

Let (2x+3y) 2=m, (3x+2y) 5=n.

then m=n+2

3m=2n+6

The solution is m=2 n=0

Bring it into the original style. 2x+3y=4

3x+2y=0

The solution is x=-8 5 y=12 5

10 on both sides of the upper and lower sides is fine, 5(2x+3y)=2(3x+2y)+2

15 (2x+3y) = 2 (3x + 2y) + 6 and then go down.

Multiply the first formula by 3,1*3-2, 2-2*1 to get 3x+2y=0,2x+3y=4,y=12 5,x=-8 5.

1 Formula 5 (2x+3y) = 2 (3x + 2y + 10) 10x+15y = 6x + 4y + 20 4x + 11y = 20 (3 formula).

2 Formula 15 (2x+3y) = 4 (3x + 2y + 15) 18x+37y = 60 (4 formula).

3 Formula 18 4 18x+11 18 4y=90 5 Formula 5-4 11 18 4y-37y=30 y=12 5

Substituting Eq. 3 yields x=-8 5

Through the points, both sides at the same time *10, about Jane is OK.

It's messy writing, make do with it! Check it out for yourself.

x=2, y=0 is the solution, then substituting the equation to get.

2m = 8

The solution yields m = 4

x = 1, y=2 are also solutions, substituting the equation gives m + 2n = 8, substituting m=4, and getting n = 2 is the inequality as.

t - 6/4 - t/2 > 1

That is, 1 2 * t > 1 + 3 2 = 5 2t > 5

So the solution set of inequalities is t > 5

Bring the two solutions of the equation into the equation separately.

There are 2m = 8 m = 4

m+2n=8 n=2

So find the solution set of the inequality t-6 m-t n 1 t-6 4-t 2>1 with respect to t

t/2-3/2>1

t-3>2

t>5

There are 2m = 8 m = 4

m+2n=8 n=2

So find the solution set of the inequality t-6 m-t n 1 t-6 4-t 2>1 with respect to t

t/2-3/2>1

t-3>2

t>5

It is obtained from the solution 1, x=2, y=0, 2m=8, m=4

From the solution 2, x=1, y=2, m+2n=8, m=4 is substituted, n=2

Then t inequality is reduced to , (t-6) 4-t 2>1, -t 4>1 2, solution, t<-2

Bring the solution of the equation into the equation.

2m+0=8

m+2n=8

The solution is m=4 n=2

Bring in inequalities.

t-6/4-t/2>1

t>5

(1)50+ (150-50)×a/10=m+(150-m)×8/10

50+100a=1200-7m

m=(-100/7)a+(1150/7)

2) [50+(1000-50) a 10] 90%m=(-100 7)a+(1150 7) substitution, sorting, get a<1280 151

a is an integer, a 8

The maximum value of a is 8

Let x years ago, then 15-x=2(9-x) 15-x=18-2x x=3 Answer: 3 years ago, if the hydrostatic velocity was x kilometers per hour, then 5(x+3) = 5x+15= 15+ x=15 Answer: The hydrostatic velocity is 15 kilometers per hour.