The Chinese Football League stipulates that each team is worth 3 points for a win and 1 point for a

Set a draw x game, because 14 rounds and undefeated, so win 14-x games, according to the win-loss relationship: 3 points for a win, 1 point for a draw, so.

x+3(14-x)=34

Solution x=4 means 4 draws.

Let x be the number of wins and y be the number of draws.

3x+y=34

x+y=14

x=10，y=4

So I drew 4 games, and I answered this question many times.

The unary equation should be 3 (14-x)+1 x=34x=4

If you draw x games, because you are undefeated, you win (14-x) games, so 3 (14-x) + 1 x = 34

8=2x, so x=4

So there were 4 draws. Look

Yes, what's the matter, it's all like this, the 2012 Chinese League 16 teams have been championed by Shanghai East Asia, the runner-up Wuhan Zall, and the relegation to the east is to be determined.

Question 1:

Let's win x games, and draw (8-1-x) games due to one loss.

3x+（8-1-x)=17

x=5 So in the first 8 games, win 5, draw 2, lose 1, get 17 points Question 2:

With 17 points in the first 8 games and 6 games to go in the future, the highest score in 14 games is 17+ (14-8) x 3 = 35 points.

Question 3: If you have scored 17 points in the first 8 games, if you have played 14 games, you will score at least 29 points

17+3x ≥29

X 4 requires at least 4 wins to score no less than 29 points.

Set the win as x, the draw as y, and the loss as z; Rule.

There is 3x+y=6

x+y+z=4 ②

x, y, z are positive integers; -2x-z=2 due to x2

When x=0, z=2 y=4 x+y+z 4 so x≠0x=1, z=0 y=3 holds.

When x=2, z=2 y=0 is not true.

Suppose there are x wins, y draws, and z losses, then there are:

3x+y=6

x+y+z=4

Case 1: Z=0, then there is 3x+y=6, x+y=4, and 2x=2 can be deduced, so x=1, y=3, z=0, reasonable.

Situation 2: Z=1, then there is 3x+y=6, x+y=3, and 2x=3 can be deduced, so x=, which is unreasonable.

Case 3: Z=2, then there is 3x+y=6, x+y=2, and 2x=4 can be deduced, so x=2, y=0, z=2, reasonable.

Case 4: Z=4, then there is 3x+y=6, x+y=0, and 2x=6 can be deduced, so x=3, which is unreasonable.

So the answer is two scenarios, 1 is a win and 1 game, and a draw is 3; Or 2 wins, 2 losses.

A tie can only score 4 points, so it doesn't hold.

Wins, 3 draws, 0 losses, exactly 3 + 1 + 1 + 1 = 6 points, establishment Wins 0 draws and 0 losses, exactly 3 + 3 + 0 + 0 = 6, the establishment of the above, 2 possibilities: 1 win, 3 draws, 0 losses or 2 wins, 0 draws, 0 losses.

Set the win as x, the draw as y, and the loss as z; Rule.

3x+y=6

x+y+z=4

x, y, z are positive integers; then x=1, y=3, z=0

1 win, 3 draws, or 2 wins, 2 losses It's a good question to think about, let's assume that the number of wins is 0 1 2, and the rest go up. . . The reason for this assumption is that 3 points are not easy to arrange, 1 point is good to arrange...

If the first 25 rounds of defeats are x, then the number of draws is 2x, then the number of wins is 25-2x-x, due to a win of 3 points, the total number of wins is (25-3x)*3, and a draw is 1 point, and the total points of the draw are 2x*1=2x. So the total integral 54 = (25-3x)*3 2x. Solve x = 3 So lose 3 games, draw 6 games, win 16 games.

It's a binary equation.

If you win x games and draw y games, you will lose 4-x-y games.

The score is 6 = 3 x + y, 0 x 4, 0 y 4 wins, 1 draw, 3 losses, 0 losses, or 2 wins, 0 draws, and 2 losses.

1. Win: 1 , Draw: 3, Loss: 0

2. Win: 2, Draw: 0, Loss: 2 (game).

If you win x games and draw y games, you will lose 4-x-y games.

x=1, y=3, z=0 or x=2, y=0, z=2

This question as a football fan can quickly tell you 1 win, 3 draws and 0 losses, or 2 wins, 0 draws and 2 losses, and only these two cases.

But to use equations, it's really not very clear what the formal steps are. You can give it a try.

Let the team win x games, draw y games, and lose z games.

x+y+z=4 （1）

3x+y=6 （2）

x, y, and z are all non-negative integers (i.e., greater than or equal to 0).

In the end, there must be two sets of solutions.

x=1, y=3, z=0 or x=2, y=0, z=2

Because xyz has to be an integer, that's a strong condition.

1. Win: 1 , Draw: 3, Loss: 0

2. Wins, 2 draws, 0 losses, 2

Win one, draw three, or win two, lose two.

Set the victory x field, remain undefeated, the loss field is 0, the draw = 11-x points: 3 * x + (11-x) * 1 = 31

2*x+11=31

2*x=20

x=10 flat=11-x=1

That is, 10 wins, 1 draw.

Solution: If x games are won, then the (11-x) games will be drawn.

The equation is: 3x+(11-x)=31

The solution is x=10

Flat: 11-x=1

10 wins, 1 draw.

What equations are you going to write about this? A total of 11 games, if you win all of them, it is 33 points, now it is 31 points, and it is only possible to win 10 and draw 1. If you win 9 and draw 2, you will only have 29 points.

Set the win field as x and the draw field as y

3x+y=31

x+y=11

Get: x=10 y=1

Solution: If x games are won, then 11 games are tied (11-x).

3x+(11-x)=31

The solution is x=10

The computer is playing well, so I should work more on the math.

1. Lose one game in the first 8 games and score 17 points. Illustrate the remaining 7 games, win 5 and draw 2.

2. There are 6 games left, and a full win is 18 points, adding up to a total of 35 points.

3. If you want to score no less than 29 points, you must achieve at least 4 wins, 2 losses or 3 wins and 3 draws in the next 6 games.

1.Win 5 games.

points, 17 + 3 * 6 = 35

3.Won 3 draws and 3 wins.

The international standard for football matches is a common one.

The combination of fractions is: 15 + 1 + 1 or 12 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1Therefore, in the first way, it is enough to win 5 games and draw 2 games.

After deducting the first 8 games, there are still 6 games, so the next 6 games win 5 games and draw one, so that the first 8 games can be drawn again. So the answer is 0

2.The highest score is less losses and fewer draws, and now it can be seen that he has drawn at least two games and lost one, then the highest score is 11 wins, 2 draws and 1 loss. That's 35 points.

4 up to 4 games.

Suppose that the number of losing games is x, then the number of draws is 2x, and the number of winning games is 8-3x

3*（8-3x）+x=17

Solve x = 1 So: lose 1 game, draw 2 games, win 5 games.

Let's get two yuan once, set to win x games, lose y games:

Then the number of draws is twice as many as the negative, i.e. 2y, according to the problem, win + draw + pay = 8, i.e. x + 2y + y = 8;

Points, win x3 + draw x1 = 17, i.e. 3x + 2y = 17;

Solve x = 5, y = 1, so 5 wins, 1 loss, 2 draws.

Assuming a win in a game, a draw in a B game, and a loss in a C game, there are:

a+2c+c=8(b=2c）

3a+2c=17

The above results in 5 wins, 2 draws and 1 loss.

Set win x, draw (14-x).

3x+1(14-x)=34

2x+14=34

x = 1014 - 10 = 4 (field).

Answer: The team has drawn a total of 4 games.

If we win x field and draw y field, then 3x+y=34, x+y=14, and solve the system of equations to obtain x=10, y=4. 10 wins, 4 draws.