Mathematical problems with binary linear equations In football league matches, three points are gained from winning one game, one point is gained from drawing one game, and zero point is gained from losing one game. A certain team has six points in four games. How many times does the team win, draw or lose?

Mathematical problems with binary linear equations In football league matches, three points are gained from winning one game, one point is gained from drawing one game, and zero point is gained from losing one game. A certain team has six points in four games. How many times does the team win, draw or lose?

Win 2, draw 0, lose 2 or win 1, draw 3, lose 0
Let a flat a field and a negative B field
The number of easy wins is not more than 2
If you win a game, a + B = 3, a + 0 * b = 6-3, get a = 3, B = 0
If you win 2 games, a + B = 2, a + 0 * b = 6-3 * 2, you get a = 0, B = 2
If win 0 field, a + B = 4, a + 0 * b = 6, a, B have no positive integer solution

1. There are 10 dimes, 5 dimes and 1 yuan coins each. 15 of them are worth 7 yuan in total. How many dimes do you take for each of them?
2. The whole journey from land a to land B is 3.3km, one uphill, one level road, and one downhill. If you keep the uphill running 3km per hour, the level road running 4km per hour, and the downhill running 5km per hour, then it takes 51 minutes from land a to land B, and 53.4 minutes from land B to land A. what are the uphill, level road, and downhill distances from land a to land B

1. Let's set 10 cents, 5 cents and 1 yuan, and take x, y and (15-x-y) coins respectively
Then: x + 5Y + 10 (15-x-y) = 70
9x+5y=80
Because x, y are integers
We get: x = 5, y = 7, 15-x-y = 3
So 5 for a dime, 7 for a dime, and 3 for a dollar
2. From a to B, the uphill is XKM, the downhill is YKM, and the level road is (3.3-x-y) km
Series of equations
x／3＋y／5＋（3.3-x-y）／5＝51／60
y／3＋x／5＋（3.3-x-y）／5＝53.4／60
The solution is x = 1.425, y = 1.725
A: from land a to land B, the uphill is 1.425 km, the level road is 0.15 km and the downhill is 1.725 km

I'm not sure whether to use equations or inequalities
A football association held a league, the points rules and reward scheme are as follows
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One win, one draw, one loss
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Integral 3 1 0
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Bonus (yuan / person) 1500 700
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At the end of 12 rounds (12 games per team), team a has 19 points
(1) How many wins, draws and losses are there for team a?
(2) If each game is one, each player will get 500 yuan. A player has participated in 12 games
The sum of his bonus and appearance fee is w (yuan)

Let team a win X Games, draw y games and lose 12-x-y games
3x+y=19 .(1)
12-x-y>=0.(2)
Then (1) is y = 19-3x, and (2) is y = 19-3x
12-x-19+3x>=0
2x>=7
x>=7/2
3x from (1)

Known point P in error! Reference source not found. Axis right, distance error! Reference source not found. Axis 3 unit length, in error! Reference source not found. Axis top, distance error! Reference source not found. Axis 4 unit length, then point P coordinate is ()
A.（-3,4）B.（3,4） C.（-4,3） D.（4,3）

To the right of the x-axis and above the y-axis are the points that show the first quadrant
Three units of length from the x-axis, so the ordinate is 3, choose D

Given a + 2B = 0, find a * a + 2ab-b * B / 2A * a + AB + b * B

Given AB = 2, A-B = 4, find (2Ab ^ 2 + 1 / (a-b) ^ 2) - (1 + 2A ^ 2B / (B-A) ^ 2)

(2ab^2+1/(a-b)^2)-(1+2a^2b/(b-a)^2)
=(2ab^2+1-1-2a^2b)/(a-b)^2
=2ab(b-a)/(b-a)^2
=2ab/(b-a)
=4/-4
=-1

For example: 3A + 2b-5a-b, where a = - 2, B = 1,

1. (3x + 2Y) + (4x + 3Y) where x = 5, y + 3
2. Simplify | 1-x + y | - | X-Y | (where x < 0, Y > 0)
3．3ab-4ab+8ab-7ab+ab X=9 Y=2
4．7x-(5x-5y)-y X=3 Y=6
5．23a3bc2-15ab2c+8abc-24a3bc2-8abcX=5 Y=7
6．2y+(-2y+5)-(3y+2) X=-3 Y=5
7.3x-[y-(2x+y)] X=-5 Y=2
8. When a = - 1, B = - 2, [a - (B-C)] - [- B - (- C-A)] =?
9. When a = - 1, B = 1, C = - 1, [B-2 (- 5A)] - (- 3B + 5C) = -______ ．
10. When 2y-x = 5, 5 (x-2y) 2-3 (- x + 2Y) - 100=______ ．
11．-5xm-xm-(-7xm)+(-3xm) X=-18 Y=9
12 3a-(2a-4b-6c)+3(-2c+2b) A=-4 B=5'
13. (5-4x) (5 + 4x) - 2x (1-3x), where x = - 2
2X - [6-2 (X-2)] where x = - 2
15. (5a + 2a2-3-4a3) - (- A + 3a3-a2), where a = - 2
16. (2m2n + 2mn2) - [2 (m2n-1) + 2mn2 + 2], where M = - 2, n = 2
17. (5a + 2a2-3-4a3) - (- A + 3a3-a2), where a = - 2
18. (2m2n + 2mn2) - [2 (m2n-1) + 2mn2 + 2], where M = - 2, n = 2
3 (AB + BC) - 3 (AB AC) - 4ac-3bc, where a = 2001 / 2002, B = 1 / 3, C = 1
20. (3xy + 10Y) + [5x - (2XY + 2y-3x)] where xy = 2, x + y = 3

2A + (5a-3b) - (a + 2b) is equal to?

2a+(5a-3b)-(a+2b)
=2a＋5a－3b－a－2b
=6a－5b

a: B = 5:3, find 5A: 2B and (2a + 3b) / (a-b)

a: B = 5:3, find 5A: 2B and (2a + 3b) / (a-b)
a:b=5:3,
Let a = 5T, then B = 3T
5a:2b=(5*5t):(2*3t)=25t:6t=25/6
(2a+3b)/(a-b)
=(2*5t+3*3t)/(5t-3t)
=(10t+9t)/2t
=19t/2t
=9.5

Calculate (5a-2b) ^ 2