Who knows how to solve a series of equations in the first class?

Let the three digits to the right of these four digits be x

7×1000+x-864=10x+7

x=681, the original four-digit number is 7681

Suppose the first alloy is 240-x grams, then the second is 240-x grams.

x = 60 should take 60 and 180 grams of kilograms for the two alloys.

Let the first two digits of these three digits be x, then the original number is 10x+55 100+x=3 (10x+5)-95

x=20, the original three-digit number is 205

Set the purchase price as a unit"1", the sell price is invariant.

1+m%)=(

100+m=

m=17

1. Set the other three digits as x

7 1000 + x - 864 = x 10 + 7x = 6812, let the first x kg, 90% x + 80% (240-x) = 240 x = 60

Question 1

Original number: 7abc

New: abc7

7 a b c

a b c 7

It follows: c-7=4 then c=1

b-c-1=6 then b=8

a-b=8 then a=6

The original number is: 7681

Question 2:

Let the first alloy be x kilograms and the second alloy be y kilograms, and the equation is:

x+y=240

x*dizzy, count, x=60 y=180

The third question is the same as the first one

The original number is ab5 and the new number is 5ab, then:

10ab+5)*3-(500+ab)=95ab=20, the original number is 205

Question 4:

Do the math yourself, but you can't do it during the exam

The first question sets the purchase price as x

x=2500

In the second question, the original planned travel time is x

20x-8=22x-22

2x=14x=7

In the third question, if B's one-hour spring crack workload is x, then A's one-hour workload is (x+2).

x+2）*5+（2x+2）*4=200

x=14

1 Set the price x x = 2500

2. Set the original planned time to x 20x-8=22(x-1) x=7 and the number of parts 20*7-8=132

3. Set A to complete x parts per hour: 5x+4*(x+x-2)=200 x=16, 16 per hour for A, 14 per hour for B.

4 Suppose the remaining part takes x hours to complete (1 20 + 1 12) x + 4 * (1 20) = 1 x = 6

5. Set up x hours of car Qilu 20* x=2

Find the balance of the problem according to the meaning of the problem and list the equations.

I leaned downstairs, and the answer was completely copied from me, is there a mistake.

1 Set the purchase price x x = 2500

2. Set the original planned time of Dahao Pei to be x 20x-8=22(x-1) x=7 and the number of parts 20*7-8=132

3. Let A complete x parts per hour: 5x+4*(roll only x+x-2)=200 x=16, 16 per hour for A, 14 per hour for B.

4 Suppose the remaining part takes x hours to complete (1 20 + 1 12) x + 4 * (1 20) = 1 x = 6

5 is set to run x hours 20* x=2

The first friend of the question set the purchase price as x

x=2500

In the third question, if B's one-hour workload is x, then A's one-hour workload is (x+2)x+2)*5+(good line 2x+2)*4=200x=14

There are x students and Y rooms in the dormitory. x, y are both positive integers x=4y+20

00<4y+20-8(y-1)<8

0<-4y+28<8

28<-4y<-20

5 so y=6

x=4×6+20=44

A: There are 44 students and 6 dormitories.

There are x dormitories.

There are four people living in each room, there are 4x people, and there are twenty people without dormitories.

So a total of 4x+20 people.

Each room lives eight people, and one dormitory is not empty or satisfied.

That is, the x-1 room is full, which is 8 (x-1) people.

The last room has 4x+20-8(x-1)=-4x+28 people, who are not empty and dissatisfied.

So 0<-4x+28<8

8<4x-28<0

20<4x<28

5x is an integer.

So x=64x+20=44

A: There are 44 students and 6 dormitories.

Set up x rooms in the dormitory, the number of people is (4x+20), and the column inequality is 0 (4x+20)-(x-1)*6 6

Solve 10 x 13

x is an integer of x=12 and x=11

When there are 12 dormitories, there are a total of 68 students.

When there are 11 dormitories, there are a total of 64 students.

There are student x dormitory rooms.

x-20=4y

y-1)8〈x〈8y

Get 5 y 7

So there are 6 rooms with 44 students.

The annual interest rate for a deposit of 200 is x, and the annual interest rate for a deposit of 3000 is y. 200x 3000y from the title

If it's solved early, why bother asking questions and giving you the answer directly.

Let B increase the temperature by one degree and save x degrees of electricity per day.

x+27）+

x=180x+27=180+27=207

Answer: After raising the temperature by one degree, B saves 180 degrees of electricity per day, and A saves 207 degrees of electricity per day, and sets A x B Y

x-y=27

x+ solved it by itself, and there is already an answer.

Let B increase the temperature by one degree and save x degrees of electricity per day.

x+27）+

x=180x+27=180+27=207

Answer: After raising the temperature by one degree, B saves 180 degrees of electricity per day, and A saves 207 degrees of electricity per day, and sets A x B Y

x-y=27x+

Because the two schools admit a total of 150 students, and the number of applicants is 54 more than 10 times this total, the total number of applicants is 150 10 + 54 = 1554.

Suppose x people sign up for school A, and there are y people who sign up for school B, according to the title:

x+y) /1 + 11%) x / 1+5%) y/(1+15%)x+y=1554

Solve the system of equations to obtain:

x = 588

y = 966

A: This year, there are 588 applicants for school A and 966 people for school B.

Assuming that each person can produce x pieces by machine and y pieces by hand every day, there are according to the topic.

x+3y=60

2x+y=80

2*Equation 1 - Equation 2, 5y=40, y=8, substituting Equation 2 to x=36

So 5x+5y = 180+40=220

So if a 5-person machine is 5 people and 5 people are hand-made, it can produce 220 pieces per day.

Hope it works, thanks

1 Set the previous annual report A has x people and the previous annual report b has y people.

A system of equations derived from the meaning of the question.

5%x+15%y=11%（x+y） （1+5%） x+（1+15%）y=150×10+54

The solution is x = 560 y=840

1+5%) x= 588 (1+15%)y=966 The number of applicants for school A this year is 588 The number of applicants for school B is 9662 If one person uses a machine to make x parts a day One person makes y parts a day by hand, and the system of equations from the problem is x+3y=60 2x+y=80 and x=36 y=8

5x+5y=5×36+5×8=220

With 5 people using machines and 5 people relying on manual work, it can make 220 pieces per day.

1. If school A recruited x people last year, school A enrolled (1+5%)x people this year, and school B recruited [(150*10+54)-(1+11%)x] (1+11%)=1440-x people last year.

So (1+5%) x + (1400-x) (1+15%) = 150*10+54

x=560 of the solution

School A enrolls 560 (1+5%) = 588 students this year.

School B enrolls 150 10 + 54-588 = 966 students this year.

2. If the machine can manufacture x pieces per day, then the manual can manufacture (60-x) 3 per day, so 2x+(60-x) 3=80

x=36 pieces are solved.

then the manual can make 8 pieces per day.

So: 5*36+5*8=220 pieces.

At 9 o'clock, there are x people at the gate of the park, and the number of people arriving at the gate of the park every minute is y

The first situation: there are 50 security channels for visitors to enter the park through the security check, visitors arrive here in a steady stream of the same number of people every minute to wait for the park, the average person through the security check channel to enter the park takes 20 seconds, until 12 noon there is no queue at the entrance of the D area, visitors can enter the park as soon as they arrive.

At this time, from 9 o'clock to 12 o'clock, the total number of new arrivals is 3 * 60 * y = 180y, and the total number of security checks in these three hours is x+180y

1 person in 20 seconds, 50 channels, 3 hours security check 50 * 3 * 60 * 60 20 = 27,000 people, that is, x+180y=27,000

The second case: from 9 o'clock to 11 o'clock, a total of security personnel, x+120y=21600

Combining the first and second cases, the two equations are solved x=10800 and y=90

In the third case, the number of visitors arriving at the entrance per minute increased by 50 to 90* people.

By 12 o'clock, the total number of people entering the park is 10800 + 135 * 60 * 3 = 35100

From 12 o'clock, visitors can enter the park through security as soon as they arrive at the entrance of Zone D.

The number of security channels is: 35100 (3*60*60 20)=65, and 15 security channels need to be added.

1.The purchase price is x yuan, 750, x = 500

2.The purchase price is x yuan, (1+150) x=1203B runs x meters, 6x=4x+300, x=150; (6+4)x=300, x=30

4.After x hours the encounter (72+96)x=210, after x=x hours the encounter x=51 84

x hours later, catch up with 96x--72x=210x=

1.Set the purchase price x 750*,05) x = 5002Set the purchase price x x = 600

3.Let B run x seconds and meet A for the first time 6x-4x=300 x=150 then B runs 600 meters.

Let B run x m x 4=(300-x) 6 x=1204 Let x hours meet after 72x+96x=210 x=Let the slow train meet after x hours of two cars 72x+96( x=Set x hours after the fast train catches up with the full car 96x-72x=210 x=

Let the age difference between the big and small elephants be x years old, then the age of the baby elephant this year is 2+x years old, and the age of the elephant this year is 14-x years old, so: x=(14-x)-(2+x), find x=4, so the age of the baby elephant this year is 2+x=2+4=6 years old, and the age of the elephant is 14-x=14-4=10 years old.