There are two math problems that I want to educate, and I want the master to solve them. Solve probl

1、a+b+c=3460

a-380)/(b+720)=5/3

b+720)/(c+c/3)=3/2

From the above equation, it is obtained: a=2380, b=480, c=6002, method 1.

If there are x students, of which 5x 11 girls and 6x 11 boys, then there are 6x 11 : 5x 11 -20 ) girls (5x 11 -20) 8 : 5x =176

Method two. From the meaning of the title, the number of boys has not changed, the original account for 8 13, now accounts for 6 11, and the number of boys is "1": then the number of boys is:

20 (11 6 13 8) 96 (person) The current number of people is: 20 (11 6 13 8) 6 11) 96 6 11) 176 (person) Wang.

Let A have a deposit of x yuan and B has a deposit of y yuan, then c has a deposit of (3460-x-y) yuan, and it can be seen from the question: (x-380):(y+720):

4 3(3460-x-y)=5:3:2, and the solution is x=2380 y=480

Question 1: A: 2380 yuan, B: 480 yuan, C: 600 yuan.

Question 2: 46

Question 1: According to the meaning of the question, the difference is 100, then the subtraction is 4*100=400, and the subtraction = difference + subtraction = 500

Question 2: Suppose the number of boxes of drinks in A and B is 5k and k, (5k-12) (k+12)=7 5

18k=144

The two cartons have a total of 6k = 144 3 = 48 cartons.

Length = 3 2000 = 6000 cm = 60 meters.

Width = 2 2000 = 4000 cm = 40 meters.

Area = 60 40 = 2400 square meters.

2400 square meters is a ratio of 1:2000 and 3 cm in length, so the actual length is 3x2000 = 6000 cm = 60 meters and the width is 2 cm x 2000 = 4000 cm = 40 meters, so the area is 60x40 = 2400 square meters.

The distance between A and B is xkm

When A walks halfway, B walks 1 3 distances, and when B walks halfway, A travels 3 4 distances (1-3 4-2 3) x = 66

x==(12/11)*66=72

The distance between A and B is 792 km

Under your known conditions, x = 36-4y-5z, so 2 (36-4y-5z) + y + 3z = 23 is simplified to y+z=7

Replace y+z=7

Bring in x+4y+5z=36 to get x+z=8

x+2y+3z=(x+z)+2(y+z)=8+2*7=22

The first question sets the distance as x

x-4 (x 12)]+x-6 (x 60)] = 66 where x 6 is the time for A to complete the journey, then half the time is 12 , 4 (x 12) is the path B took BC x-4 (x 12) is AC The rest is similar, omitted.

As a result, please ask for yourself.

The second question order. a=x+4y+5z=36

b=2x+y+3z=23

a+2b ： 4a+b = 1：2a：b=3：2

Then ·2a+3b=7x+14y+21z=152x+2y+3z=(7x+14y+21z) 7=22

Let's look at the second question first:

x+4y+5z=36

2x+8y+10z=72 2x+y+3z=23 subtracted to obtain:

7y+7z=49

y+z=72y+2z=14 x+4y+5z=36 subtract the two formulas to obtain: x+2y+3z=22 Question 1:

Let A and B take x and y hours respectively to complete the journey.

The first equation: the distance is the same, so 6x=4y

The second equation:

ac+cb=ab(1)

ad+db=ab(2)

and cb = 4 * x 2

ad=6*y/2

1) + (2) can be obtained:

66+4*x/2+6*y/2=2*6x

Two equations can be solved in one piece.

If you don't understand, you can ask

Under your known conditions, x = 36-4y-5z, so, 2(36-4y-5z) + y+3z = 23, simplified to 7(y+z) = 39

Then x+2y+3z=36-4y-5z+2y+3z=36-2(y+z)=36-2*(39 7).

It's just non-stop substitution, it's very simple.