It is known that the linear function is parallel to the line y = 2x / 3 + 1, and intersects with the line x-2y-4 = 0 at the same point of X axis,

It is known that the linear function is parallel to the line y = 2x / 3 + 1, and intersects with the line x-2y-4 = 0 at the same point of X axis,

Take y = 0 into x-2y-4 = 0
x=4
Intersection (4.0)
y=2/3*x+b
Bring in (4,0)
b=-8/3

2x+y-2=0,x-3y-1

2X + Y-2 = 0, x-2y + 4 = 0 below, x-3y-1 = 0 above
X ^ 2 + y ^ 2 is the square of the distance to the origin
And the farthest to the origin
It's obviously the intersection of x-2y + 4 = 0 and x-3y-1 = 0 (- 14, - 5)
x+y=-19
(Y-2) / (X-2) is the slope of the line passing (2,2)
The intersection of the three lines is (0,2), (1,0), (- 14, - 5)
So over (1,0) k is the largest and over (0,2) k is the smallest

The analytic expression of F (x) can be obtained by crossing the origin of the image of the first-order function f (x) and the intersection of the line 3x + y = 5 and x-2y = - 3

The intersection point of the straight line 3x + y = 5 and x-2y = - 3 is required, and the equations of two straight lines are composed of equations, that is
（1）3x+y=5
（2）x-2y=-3
The solution of the equations is: x = 1, y = 2, that is, the intersection is (1,2);
Let f (x) = ax + B
The origin (0,0) and intersection (1,2) are brought into the function, and the solution is obtained
a=2 b=0
So the analytic expression of the first-order function f (x) is: F (x) = 2x

The image of the first-order function y = KX + B is parallel to the line y = (1-3x) / 2, and intersects with the line 3x-2y-1 = 0 at the same point on the y-axis

∵ the image of the first-order function y = KX + B is parallel to the line y = (1-3x) / 2 ∵ the analytic expression of the first-order function can be: y = (- 3 / 2) x + B ∵ the same point on the y-axis where the image of the first-order function intersects with the line 3x-2y-1 = 0 ∵ let y = 0,3x-0-1 = 0, x = 1 / 3 ∵ the intersection of the image of the first-order function and the y-axis (0,1 / 3)

Given that the solution of the equations 3x + 4Y = 6,2x-3y = m is x = 2, y = 3, then the linear function y = 3 / 4 + 3 / 2 and
Y = 2 / 3-1 / 3 of the image intersection coordinates are

Sorry! The original problem is wrong: x = 2, y = 3 is not the solution of the equation 3x + 4Y = 6. I don't know whether the equation is wrong or the solution is wrong, so I can't solve the problem directly. The solution is as follows: ① substitute the solution of the original equation into the equation containing m, solve an equation about M, and get the value of M; ② substitute the value of m into the analytic expression of the first-order function containing m, and regard it as

Solve the equations {X-1 of 3 = Y-3 of 4 = Z of 5 + 2 {2x-3y + 2Z = 1}

（x-1)/3=（y-3)/4=(z+2)/5 ①
2x-3y+2z=1 ②
Equation 1 × 60: 20x-20 = 15y-45 = 12z + 24
x=(12z+44)/20=(3z+11)/5 y=(12z+69)/15=(4z+23)/5
In the second generation of x = (3Z + 11) / 5, y = (4Z + 23) / 5, we get the following result:
2(3z+11)/5-3(4z+23)5+2z=1 (6z+22-12z-69)/5+2z=1 -6z-47+10z=5 4z=52 z=13
∴ x=10 y=15 z=13

Given - [- (- a)] = 2, find the opposite number of A. if A-2 and - 7 are opposite numbers, find the value of A. if x and y are opposite numbers, find the value of X + y in 2010

-[-(-a)]=2=-a
The opposite number of a = - a = 2
a-2=-(-7)
a=2+7=9
The sum of two opposite numbers is 0, so x + y = 0
2010 x + y = 2010 0 = 0

The product of all integers whose absolute value is less than 2014 is

0 and 0 are also integers, so any number multiplied by 0 gets 0

What is the product of all integers whose absolute value is less than 2014?

It contains zero, so it's zero

The sum of all integers whose absolute value is not greater than 2014 is______ ．

All integers with absolute value not greater than 2014 are - 20142014, - 20132013 , - 2, 2, - 1, 1, 0, so according to the sum of two opposite numbers is 0, the sum of these numbers is 0, so the answer is: 0