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If the ten digits are x, then the single digit is 2x

So the original number is 10x+2x

After modulation, it is 10*2x+x

So (10*2x+x)-(10x+2x)=3621x-12x=36

9x=36x=4,2x=8

So these two numbers are 84 and 48

Let the tens be: x

Then x * 10 + 2x + 36 = (2x)*10 + x12x +36 = 21x

9x = 36

x = 4 gives the number 1 = 48 and the number 2 = 84

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If the ten digits are x, then the single digit is 2x

So the original number is 10x+2x

After modulation, it is 10*2x+x

So (10*2x+x)-(10x+2x)=3621x-12x=36

9x=36x=4,2x=8

So these two numbers are 84 and 48

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Let the single digit be 2x and the ten digit be x

10x+2x)-(20x+x)=-36

x=4 so these two numbers are 48 and 84

Let the ten digits be x, then the single digit is 2x, and the question is: (2x*10+x)-(x*10+2x) 36

x=4 2x=8 i.e. the original number is 48

Let this number be (ab).

Then according to the condition b=2a

10a+b)-(10b+a)=36

Solving these two equations gives us a=4 b=8

So 48 84

In the first problem, first substitute x=-2 into the equation (k-2)*-2=-2-1 solution -2k+4=-3

2k=-7k=The second problem can be solved according to the meaning of the equation 8x-7+6-2x=0, and the solution is 6x=1

x=1 6, the third question, x=-3 2, substituting into the equation, can get 4(-3 2-3a) = 1-7a solution, -6-12a=1-7a

I don't know how to add -5a=7 a=-7 5

Solution: There are x rabbits.

4x+2（23-x）=62

2x+46 =62

2x =62-46

x=16÷2

x=823-8=15.

A: There are 15 chickens and 8 rabbits.

1. Chickens and rabbits in the same cage. There are 23 heads with 62 legs. How many rabbits are there? (Solve with equations) set rabbits x only.

4x+2(23-x)=62

4x+46-2x=62

2x=16x=8

If the chicken is x, the rabbit is 23-x

According to the title (because there are 2 chicken legs and 4 rabbit legs).

2x+4(23-x)=62

x=15A: 15 chickens and 8 rabbits.

A car travels at a speed of 40 kilometers per hour from A to B, and after driving for 3 hours, the average speed is forced to decrease by 10 kilometers per hour due to rain, and the result is that it arrives at B 45 minutes later than expected.

Solution: Because of the rain, the average speed was forced to decrease by 10 kilometers per hour, and as a result, it was 45 minutes later than expected.

That is, 45 minutes = 3 4 hours, which is equivalent to 40 3 4 = 30 (kilometers) less

30 10 = 3 (hours) That is, the average speed is forced to decrease by 10 kilometers per hour due to rain, and the time it takes to travel the distance.

Therefore, the distance between A and B is 40 3 + (40-10) 3 = 120 + 90 = 210 (km).

Solve with equations: Solution: Let the estimated time be x. According to the inscription column.

40x=40×3+（40-10）×(x-3+3/4)

40x=120+30x-90+90/4

10x=210/4

x=21/4

That is, the distance between A and B is 40 21 4 = 210 (km).

A: The distance between A and B is 210 km.

The distance between A and B is S km.

s/40 = 3+ (s-40*3)/(40-10)-45/60s/40=3+

s=120+1200s-16000-30

1199s=15910

s=15910/1199

This kind of problem is still an equation. It takes x hours to reach location B under normal conditions.

Then the distance between the two = 40x = 40x3 + 30 (x + 45 minutes is the hour.)

If we get 10x=, then 40x=210 is the distance between the two.

If x mothers have lost 2 children, then there are (40-x) mothers who have lost 1 child.

You can list equations:

2x+(40-x)=70

x=30, so there are 30 mothers who have lost two children.

1.Solution: Let the side door pass x people per minute on average, and the main door can pass x + 40 people 2 (x + 40 + x) = 400

2(2x+40)=400

2x+40=200

2x=160

x=80x+40=80+40=120

A: 120 people can pass through the main gate and 80 people can pass through the side door every minute2Solution: It takes x minutes to pass through.

120×2+80)*（1-20%）x=45*4*6256x=1080

x=so comply with safety regulations.

Set up the main gate to pass x students per minute, you can get x+x-40=200, and solve x=120

Therefore, the main gate can pass 120 students per minute, the side gate can pass 80 students per minute, 3 doors can pass 120 + 120 + 80 = 320 students per minute, after the efficiency is reduced, it can pass 320 * (1-20%) = 256 students per minute, and 5 minutes can pass 256 * 5 = 1280 students.

The total number of students is 45*6*4=1080

So it meets safety regulations.

How many students can pass through one front gate and one side gate per minute?

Set up the main gate is divided into x people per minute, and the side gate y people have:

2（x+y）=400:

x-y=40;

x=120, y=80;

In an emergency, due to the crowding of students, the efficiency of going out is reduced by 20% x = , y = 5 (2x+y) = 1280 students in the whole building at most: 4 6 45 1080 people do not comply with safety regulations, because 1280 1080 means that all can leave in the worst case.

(1) Set up a side door to pass x students per minute.

2x+2（x+40）=400

The solution is x=80 x+40=120

So 120 people pass through the main gate every minute, and 80 people pass through the side gate every minute.

2) Number of students = 4 * 6 * 45 = 1080 students.

5 (80 + 120 * 2) * (1-20%) = 1280 1080 so in line with safety regulations.

There is one main gate every minute that can pass through x students.

x+（x-40）】*2=400

x = 120 people – per minute at the main entrance.

120-40 = 80 people – side door per minute.

The maximum capacity of the whole building is 45*4*6=1080 people.

Evacuation in 5 minutes in case of emergency:

120*2+80)*(1-20%)*5=1280 people More than 1080, so it meets safety regulations.