To solve with a binary equation, the process must be written

Question 2. 2x-7y=3 1 formula.

4x+2y=-2 2 formula.

Equation 1*2 yields 4x-14y=6 3.

Eq. 2-3 gives 16y=-8 and gets y= -1 2 Substituting Eq. 2 Solution gives x= -1 4

Question 3. 3y-4x=6 gives this equation *3 to 9y-12x=18 1.

3x-4y=-1 gives this equation *4 to 12x-16y=-4 2.

Formula 1 + 2 gives -7y=14 y= -2

Substituting y=-2 into 3y-4x=6 solves the fourth problem of x=-3. 3 x + 1 + 2 y-1 = 2 1.

3 2x-1 + 2 1-y = 1 2.

1 + 2 3 x+1 + 3 2x-1 = 3 pass points multiply (x+1) (2x-1) on the left and right sides.

3(2x-1)+3(x+1)=3(x+1)(2x-1) divided by 3 to get 2x-1+x+1=(x+1)(2x-1)3x=2x 2+x-1

2x^2-2x-1=0

The solution gives x1 = 1 2 (1 - root 3) x2 = 1 2 (1 + root 3) corresponding to y1 = 2 - root 3 y2 = 2 + root 3

This kind of question is also taken out.,Speechless.。。。

I'll tell you how to do it:

Treat one of the unknowns as known to represent the value of the other.

Substituting it into another equation can find the value of an unknown.

The other came out naturally.

Set up large boxes and small boxes, each box contains x and y bottles;

3x+4y=108 (1)

2x+3y=76 (2)

1) 2-(2) 3:

8-9)y=216-228

y=12 brings y into (2) to get: x=20

A: .Large and small boxes contain 20 and 12 bottles each?

Set up large and small boxes of x and y bottles.

3x+4y=108 ..1)

2x+3y=76 ..2)

2(1): 6x+8y=216 ..3)3(2): 6x+9y=228 ..4)(4)-(3): y=12

Substitution (2): 2x+36=76, 2x=40, x=20 so 20 bottles in a large box and 12 bottles in a small box.

Solution: Set up a large box x bottle, a small box y bottle.

3x+4y=108（1）

2x+3y=76（2）

2) 3-(1) 2.

y=76×3-108×2

y=12 substitute y=12 into (1).

3x+48=108

3x=60x=20

A: 20 bottles in a large box and 12 bottles in a small box.

Let the large box be x and the small box be y

3x+4y=108 ..1)

2x+3y=76 ..2) (1)*2 -(2)*3.

y=228-216=12

x=20

Set up large boxes, each box contains x small boxes, and each box contains y

The system of equations 3x+4y=108 2x+3y=76 is obtained

We get x=20 y=12

Let x,y

3x+4y=108

2x+3y=76

The simultaneous solution is x=20 y=12

That is, 20 boxes of large boxes and 12 boxes of small boxes.

Hope it helps.

Set up large box x bottle, small box y bottle.

3x+4y=108

2x+3y=76

Solution: (3x+4y)-(2x+3y)=108-763x+4y-2x-3y=31

x+y=31

3x+4y=108

3(x+y)+y=108

3*31+y=108

93+y=108

y=108-93y=15

Set up large box x bottle, small box y bottle.

3x+4y=108 ①

2x+3y=76 ②

2 *3=6x+8y-(6x-9y)=108*2-76*3So:y=12

x=20

Set: The big box has x bottles per box, and the small box has Y bottles per box

3x+4y=108

2x+3y=76

Solution: x=20 y=12

Large box x small box y 3x+4y=108 (1) 2x+3y=76 (2) (1) Multiply 2 3 on both sides to get 2x+8 3y=72(3) (2)-(3)1 3y=4 y=12 Bring in (1) to get x=20

1) Let Mr. Li surf the Internet for more than B hours for Zheng Chai, which can be obtained from the title, a+b=80, b=80-a

a 4 + b 8) = 736-80 60 , substituting b = 80-a to obtain.

4a+8 (80-a)=384, 4a=256, a=64so, b=80-64=16

The prescribed time A of the city was 64 hours, and Mr. Li spent more than 16 hours on the Internet.

2) The average monthly cost of using broadband for half a year:

Without broadband, 80 hours of Internet access per month, the cost of 736 is required, so it is appropriate to change to a broadband bureau.

Untie. 1) 4a+(80-a)x8=736-(80x60) 640-4a=384

a=64Answer: A is 64 hours, and Li Xianshu has been hidden for more than 16 hours.

2) He is suitable for broadband Internet access.

Solution: The time for Mr. Li to surf the Internet for more than a is b hours, which can be obtained from the title, a+b=80, b=80-a

a 4 + b 8) = 736-80 60 , substituting b = 80-a to obtain.

4a+8 (80-a)=384, 4a=256, a=64so, b=80-64=16

The city's time limit is 64 hours, and Mr. Li has been online for more than 16 hours.

2) The average monthly cost of using broadband for half a year:

Without broadband, 80 hours a month on the hail lack of network, the cost of Li Jing is 736 Answer: It is appropriate to change to broadband.

1. Substitution method.

Use y to represent x, take your question above.

8y=2x, and bring 2x into 3y=2x+4.

y=-4/5

2. Addition and subtraction.

Find the least common multiple of the coefficient.

Multiply 3y=2x+4 by 8 at the same time

Multiply 8y=2x by 3 when both sides are on the same slide

That is, the credit is worthy of 24y=16x+4

24y=6x

Subtract the new two formulas.

You can get x=-2 5

Set Bi He Tong x,y

3x+4y=108

2x+3y=76

Join forces to solve x=20

y=12 is 20 boxes of large boxes.

Box of 12.

I hope I can help you.

Question 2. 2x-7y=3

Type 1. 4x+2y=-2

Type 2. Equation 1*2 yields 4x-14y=6

Type 3. 2-3 formula.

16y=-8 gets y=

Substituting 2 formulas. The solution is x=

Question 3. 3y-4x=6

Equirety*3 is obtained.

9y-12x=18

Type 1. 3x-4y=-1

This equation *4 is obtained.

12x-16y=-4

Type 2. 1 + 2.

Get. 7y=14y=

Substitute y=-2 into the bend talk 3y-4x=6. Solve x = the fourth question. 3/x+12/y-1

Type 1. 3/2x-12/1-y

Type 2. 1 + 2.

3 x+1

3/2x-1

Scores. Multiply (x+1) (2x-1) on the left and right sides.

3 (2x-1) + 3 (x+1) = 3 (x+1) (2x-1) divided by 3 to get Jianzen.

2x-1+x+1=(x+1)(2x-1)

3x=2x^2+x-1

2x^2-2x-1=0

The solution is x1=1 and the burial is 2 (1-3).

x2 = 1 2 (1 + root 3).

Corresponding to y1 = 2 - root 3

y2 = 2 + root 3

Let the number of tens be x, then the single digit is x+5

The value of this number can be expressed as 10x+x+5=11x+5, and now the value of the new number becomes 10(x+5)+x=11x+50, and the sum of the two numbers is 22x+55=143, 22x=88, x=4, and the original number is 11*x+5=49