and differential multiplier problems production, workmanship, and other problems .

Travel issues. 2.A velocity x B velocity y

then 40x+90x+90y=18

108x+108y=18

3.Let the x-kilometer equation be set.

x/5.First, find out the sum of the speeds of the two cars is (200+280) 16=30, and know the speed ratio of 3:2

So 30 (3+2)=6

Bus speed 6 * 3 = 18

Train speed 6*2=12

It's a matter of time. Another 8/11, this is a typical pursuit problem, the hour hand is in front and the minute hand is behind, according to the speed and "distance" known, it is natural to find, here is not a detailed calculation, I have seen this question, so I will give you the answer directly.

Boating problems. 12.The speed of the ship v

2v+6=3v-9

So v=15

Then the distance is 2v+6=36

13.Set the aircraft speed v

17(v+24)/6=3(v-24)

v=840, then the distance is 3(v-24)=2448 water pipe problem.

3) At the same time turn on the water intake per hour 1 15-1 24

4) Time is 13 15x where x is the answer to the third question.

and differential multiplier problems.

1.One person does 1 40 in 1 hour, first let x people do it for 4 hours, and then x+2 people do it together for 8 hours.

So x 10+(x+2) 5=1 x=2

Match points issues.

Let the original length xx 2+1) 2-1=

The solution is x=12

Solve zero problems professionally, just think it's good, and the score is just a passing moment.

Set: A can make x parts per day. Then B can make x+1 parts per day.

According to the title, the following equation holds:

30/x=30/（x+1）+

Solve the equation: x=4

Answer: Worker A can make 4 parts a day, and the original plan is to complete it in one day.

Solution: If worker A makes x parts per day, then worker B makes (x+1) parts per day. If the original plan is to be completed in Y days, there will be.

x*y=30---1)(x+1)*(

Solve the system of equations composed of (1) and (2) to obtain.

x=pcs) y=4 (days).

After testing, it fits the topic!

Assuming that the original plan is completed in x days, A will make y parts per day;

From the title: 30=x*y.

30=(y+1)*(Eq. xy+ Eq.

formula substitution.

x= and xy=30 x=30 y get y +y-20=0 (x, y>0).

The solution yields y=4 and yields x=

A: The original plan was to be completed in one day, and A made 4 parts a day.

A does 4 a day, originally planned for a day.

The sum problem is a problem of finding out how much each of these two numbers is when the sum of two numbers is known and the relationship between the multiples of two numbers.

The difference multiple problem is a problem that knows the relationship between the difference between two numbers and the multiple of two numbers, and finds what these two numbers are.

And times the problem solving ideas:

Simple questions use formulas directly, and complex problems use formulas when they are adapted.

Formula: Two Numbers and Parts Sum = Decimal.

Decimal multiples = large numbers or two numbers and - decimal numbers = large numbers.

Difference Factor Problem Formula:

Difference (multiple-1) = decimal place;

Decimal + Difference or Decimal Multiples = Large Numbers.

Sum multiple problem: Knowing the relationship between the sum of two numbers and the multiple of two numbers, we usually call it the sum multiple problem. When solving such application problems, it is necessary to draw a line segment diagram according to the conditions and problems given in the question, so that the quantitative relationship is clear at a glance, so as to find out the law of solving the problem and solve it correctly and quickly.

Formula: Two Numbers and Parts Sum = Decimal Decimal Multiples = Large Number or Two Numbers and Decimal = Large Number.

Suppose he wholesaled x kg of cucumbers, then he wholesaled 40-x kg of eggplant, and the solution was x = 30 kg.

The amount of money earned: 30 * (RMB.

Set to buy cucumbers x kilograms; x*；x=10；So bought eggplant is 30 kg; The money earned is; 10*（

A hen lays an egg, a duck lays three eggs, a chicken and a duck have a total of ten, a total of twenty-two eggs, how many chickens and ducks are there?

Let B produce x per day.

A is (3 4) x C is (3 2) x

So (3 4)x+(3 2)x=945+x gives x=756

B is 756 and A is 567

C is 1134

Solution: Let B produce x per day.

A is (3 4) x C is (3 2) x

So (3 4)x+(3 2)x=945+x gives x=756

A: 756 cases for B, 567 cases for A, and 1,134 cases for C.

1.Let the second name be x

then (x+250)+(x)+(x-125)=8753x=875+125-250

3x=1000-250

3x=750

x=2502.Apple trees have 1800 (1+2+3)=300 and oranges have 300x3=900

The pear tree has 300x2=600

3.A180 (2+2x3+1)=20

B 20x2=40

C 40x3 = 120

4.Set up x tons of armor and hidden meat.

x+3x-4=92

4x=96x=24

Otozo meat is 96-x=96-24=72

4.A did (160-6-7) (1+2).

B did 160-49=111

4x40=160

2.Set the original investment of x yuan.

x+25=3(x-3)

x+25=3x-9

34=2xx=17

3.Let B be x

3x+2-x=50

2x=48x=24

4.Let's say there are x kilograms of potatoes each.

x-14=3(x-38)

x-14=3x-114

2x=100

x=505.According to the first condition, the first bookshelf has 14 more books than the second bookshelf, and when the first bookshelf has 14+2 12=24 books more than the second bookshelf, the number of books on the first bookshelf is twice the number of books on the second bookshelf, so the number of books on the second bookshelf is 38 books at this time, plus the original 12 books, so there are 50 books on the second bookshelf and 64 books on the first bookshelf.

2x2500-3000=2000

b 2000÷1=2000

a 2000x3=6000

And is the addition of two numbers, for example: A makes 10 parts, B makes 12 parts, how many parts does A and B make in total?

10 + 12 = 22 (pcs).

The difference is the subtraction of two numbers, for example: A makes 10 parts, B makes 12 parts, how many parts does A make less than B?

12 = 10 = 2 (pcs).

The multiple is multiplied, for example: A makes 10 parts, B makes 2 times the number of parts, how many parts does B make?

10 2=20 (pcs).

1. The number of balloons in pile A and B and C is more than 5 in B pile: the original C pile is 52 than A pile, A is more than 4 balloons in B and B is given to C and 5 balloons are known: C is 143 than A and C is 143 than A and 6 balloons are known as C and A is 2 in C and A is 2

4. The number of balloons in pile C is 2 times less than that of pile A, 20 times less than pile C, and 20 times less than pile C, so I add 20 to pile C (pile C is 22 more than pile A) and pile C is just 2 times that of pile A: 22 (2-1)=22 ().

Suppose last year's income is x yuan, and the expenditure is y yuan, and the system of equations is obtained

x-y=1200

1+15%)x-(1+5%)y=1200+1140 will be simplified to:

23x-21y=46800

Solve this system of equations to get:

x=10800, y=10800-1200=9600, so last year's income was 10,800 yuan, and the expenditure was 9,600 yuan.