(y-3)/4=x，6(x- 1)+2 & gt； = Y. y=4x+3 is obtained from the equation and substituted into the following inequality: 3 >; =3.5。

Take x=4, the test is correct, y= 19.

If x=5 (or greater), let's satisfy the problem.

2. The length and width of the football field are measured in meters. Column inequality:

2(x+70)>350，70x & lt； 7560。 Solution: 105

According to the conditions given in the title, this football field can be used for international competitions. (The title condition "Length between 100m and 100m" is incorrect.

3. "If the previous 13 class is divided into 13 sets" should be changed to "If the previous class is divided into 13 sets".

Number of classes x, number of Fuwa sets y. According to the meaning of the question:

y= 10x+5， 13(x- 1)+4 & gt； Y. substitute y= 10x+5 into inequality:13x-9 > 10x+5, which is 3x >；; 14，x & gt=3。

With x=3 test, does not meet; Take x=4, which is inconsistent; X=5, y=55 is more appropriate. If x=6, it is obviously unreasonable that the last class can't get Fuwa. So x=5, y=55 is the solution of the problem.

4、x+y=-2...( 1)，5x-3y=2a...(2)。

( 1)* 3+(2):8x = 2a-6 & lt； 0, get a

( 1)* 5-(2):8y =- 10-2a & lt； 0, get a & gt-5.

So a should be -4, -3, -2,-1, 0, 1 or 2.

5,2x+a > 3,5x-b < 2, x >(3-a)/2, x & lt(2+b)/5. According to "the solution set is-1 < x < 1", then

(3-a)/2=- 1，a=5，(2+b)/5= 1，b=3。

Therefore, (a-2b) 2010 = (-1) 2010 =1.

6. The students in Class A and Class B are X and Y respectively.

300 & lt6+9(x- 1)= 13+8(y- 1)& lt； 400。 Simplified, 300

303 & lt9x & lt403，295 & lt； 8y & lt395。 The number divisible by both 8 and 9 in this range, that is, the multiple of 72, is only 360. So 9x=360 and x=40. Substituting 9x-3=8y+5 gives 8y+5=357 and y=44.

So the total number of students in Class A and Class B is 84.