Use a system of binary equations to solve the process

Solution: If A's income is x yuan and the expenditure is y yuan, then B's income is 4x 5 yuan and the expenditure is 2y 3 yuan.

x-y=1500 1)

4x/5-2y/3=1500 2)

1)-2), de.

x/5=y/3

x=5y/3

Substituting 1), get.

2y/3=1500

y=2250 yuan.

x=5*2250 3=3750 yuan.

4x 5 = 3000 RMB.

Answer: A's income is 3,750 yuan, and B's income is 3,000 yuan.

Set A's income x and monthly expenditure y

x-y=（4/5）x-（2/3）y=1500

Solution x=3750 B income 3750 5*4=3000

Let A's approximate income be A and B's B

then there is a:b=5:4

a-1500）：（b-1500）=3：2

Just solve the problem. a=3750

b=3000

Solution: Let A's income be x yuan and B's income be y yuan.

x:y=5：4

x-1500):(y-1500)=3:2 from x=5 4y, bring in (5 4y-1500):(y-1500)=3:2

Using the basic property of proportion, the product of the inner term is equal to the product of the outer term: 2 (5 4y-1500) = 3 (y-1500).

Then solve this unary equation on it, put the parentheses.

2×5／4y-2×1500=3×y-3×15005/2y-3000=3y-4500

3y-5/2y=4500-3000

1/2y=1500

y=1500÷1/2

y=3000

Substituting y=3000 into x:3000=5:4, still using the product of the inner term is equal to the product of the outer term, 4x=3000 5, we get x=3750

Answer: The income of A and B is 3,750 yuan and 3,000 yuan respectively.

1.The weight of salt in brine with 10% salt content + the weight of salt in salt water with 85% salt content = the weight of salt salt in salt water with 45% salt content.

The columnable equation is: 10%x+85%y = 45%*12

2. The weight of brine containing 10% of salt + the weight of brine containing 85% of salt = the weight of brine containing 45% of salt.

The columnable equation is: x+y=12

1. The total sales price of candy sold per kilogram of blue + the total sales price of sugar to dust fruit sold per kilogram = 3,The total price of candy sales is 6 yuan.

The columnable equation is:

2. The weight of candy sold per kilogram + the weight of candy sold per kilogram = the weight of candy sold per kilogram.

The columnable equation is: x+y=200

Juvenile, I seriously suspect that you copied the wrong question, and you show us the question, and you don't need to list the process.

Do not write) to the two equations to label He Quarrel 1 and Zen Pants 2

Solution: Give 1 by 12 and get.

8(x-y)-3x-3y=-1

8x-8y-3x-3y=-1

5x-11y=-1 3 (pure enlightenment).

Obtained by 2 formulas. 3x+3y-2x+2y=3

x-5y=3

x=3+5y4(formula).

Substitute 4 into 3.

5（3+5y)-11y=-1

15+25y-11y=-1

14y=-1-1x5

y=-16\14

x=3+5y

x=42\14+(-80\14)

x=-38\14

x=-2+5\7

3x-y+z=3 (1)

2x+y-3z=11 (2)

x+y+z=12 (3)

1)-2), de.

5x-2z=14

z=(5x-14)/2

2)-(3), got.

x-4z=-1

z=(x+1)/4

5x-14) 2=(x+1) collapse 4

I tidy it up and get it. 9x=29

x=29 is next to a circle of 9

z=(x+1)/4=19/18

x=29 9 z=19 18 substitution (3).

y=12-x-z=12-29 9-19 18=139 18x=29 9 y=139 18 z=19 do 18