In the first year of junior high school, let s use a system of binary equations to solve it.

Assuming the sum of the ages of the children this year is y, the couple has x children in total.

Then the sum of the couple's ages this year is 6y.

Two years ago (everyone's age is subtracted by two years):

The sum of the couple's ages is (6y-2*2)=6y-4; The sum of the ages of their children is (y-2x); then there is 6y-4=10*(y-2x);

After six years (add 6 to everyone's age):

The sum of the couple's ages is 6y+2*6=6y+12;The sum of the ages of their children is (y+6x); then there is 6y+12=3*(y+6x);

There are two sets of equations:

6y-4=10*(y-2x)；

6y+12=3*(y+6x)；

Solution x=3

How many children does a couple have when they are six times the age of their children, who are 10 times the age of their children two years ago, and who are three times the age of their children six years later?

Test Topic: Application of Ternary Linear Equations

Analysis: Let the sum of the current ages of the couple be x, and the sum of the ages of the children be y, and there will be n children in total, and the system of equations about x, y, and n is established

Answer: Solution: Suppose the sum of the current ages of the couple is x and the sum of the ages of the children is y, and there are n children in total, and the sum of the current ages of the couple is 6 times the sum of the ages of their children

x=6y, which is 10 times the sum of their children's ages two years ago: x-2 2=10 (y-2n), which is 3 times the sum of their ages after 6 years and 3 times the children's ages 6 years later: x 2 6=3 (y 6n), and the system of equations is listed.

x-2×2=10×(y-2n)

x 2×6=3×(y 6n)

Substituting x=6y into the system of equations.

Solution: n=3

A: The couple has three children together

Good luck with your studies.

Three-dimensional bar. Solution: Suppose the sum of the ages of the couple is x and the sum of the ages of the children is y, and the couple has z children.

x=6y, x-4=10(y-2z), x+12=3(y+6z) gives {x=84, y=14, z=3

A: The couple has three children.

There are x children, and the sum of the children's ages is y

6y-2×2=10(y-2x)

6y+6×2=3(y+6x)

Solution x=3

1.There are x questions for difficult questions, y questions for easy questions, and z questions for medium questions. According to the meaning of the question, the system of equations is obtained.

x+y+z=100

x+3y+2z=180

*2-, i.e., 2x+2y+2z-(x+3y+2z).

x-y=20，2.Among the 100 questions, there are x difficult questions and y simple questions. then the medium title is (100-x-y).

According to the title:

x+2(100-x-y)+3y=60×3

Because only one person can solve the difficult problem, two people can solve the medium problem, and all three people can solve the easy problem. Three people solve a total of 60 3 problems, and repetitions are counted).

Solve the equation and get.

y-x=-20

i.e. x-y=20

x-y = difficult – easy = 20

There are 20 more difficult questions than easy questions.

That's how it should be. I don't know if it's right or not...

I don't seem to be able to express myself very clearly, so I don't know if I understand it

1.No binary equations.

It can be obtained according to the title.

The solution is m=8, and 12-8=4.

2.Set Xiao Ming to vote for x, Dad for y, get.

x+y=20

3x=y gives x=5

y=15

Solution: Let a student's average score in the first four times be x points, and the fifth test score is y points, according to the meaning of the question

4x+y) 5 = 90, i.e. y = 450—4x.

Obviously, the score of the fifth test is higher than 90 and not more than 100, so there are 90 < 450-4x 100.

Solution: x <90.

x is an integer.

x = 88 or 89.

This is due to the fact that the scores of the two students are different, i.e. the average score of the first four times is 88 points and the other is 89 points. So the fifth time the score was 98 points and the other was 94 points.

The average score of the first four tests is X, and the score of the fifth test is Y

then there is (4x+y) 5=90

y=450-4x

And there are 90 < y 100, which is 90 < 450—4x 100 to get x <90.

x is an integer less than 90 points, then x

So the average score of these two students is 88 for one and 89 for the other

1.Set volleyball xThen (2x-3):

x=3:2 gives x=6Platoon = 6 Baskets = 2x-3 = 9.

2.Set the uncle is now age x, Xiaoqiang ythen x=2y-4 (x-y)+x=40

Uncle De x = 28 years old, Xiao Qiang y = 16 years old. 3.Set x box body and Y box bottom.

Then x+y=36 25x 40y=1 2 get the box body x=16Bottom of the box y = let 30% be x and 75% be y. then 30%x+75%y=50%*18 x+y=18

then we get x=10kg, y=8kg 5.Set the unit price of toothbrush x

Toothpaste is yThen according to the known conditions, the equation is 39x+21y=396 and both sides of the whole formula are divided by three at the same time, and the equation becomes 13x+7y=132Then multiply both sides of the equation by four at the same time, and the equation becomes 52x+28y=528

So the account of 52x+28y=518 is wrong. Thanks for the reference! Hope it helps!

1. Set the number of basketballs to x volleyball book y

2y-3=x,3y=2x

Substituting y=6,x=9

2. Set Xiaoqiang's age to be y, and Uncle X's is Y

x-4=y-x,y-x=40-y

So 12x+4-12y+6=180

12x-9-12y-12=12

So 12x-12y=170

12x-12y=33

So there is no solution.

Original formula = 12x + 4-12y + 6 = 180 12x-9-12y-12 = 1 substitution elimination = 170-12y) 12 + 12y = 2285 6-y + 12y = 22

85/6+11y=22

11y=132/85

y=。。Finally, uh, it's a lot of numbers.