If the positive integer solution of inequality 3x-k less than or equal to 0 about X is 1.2.3., what is the value range of K

If the positive integer solution of inequality 3x-k less than or equal to 0 about X is 1.2.3., what is the value range of K

K is greater than or equal to 3x, that is, K is greater than or equal to 3 and less than or equal to 9

If the positive integer solution of inequality 3x-a ≤ 0 is 1, 2, 3, then the value range of a is __

Solve the inequality 3x-a ≤ 0 and get x ≤ a
3,
∵ the positive integer solution of the inequality is 1, 2, 3,
∴3≤a
3<4,
The solution is 9 ≤ a < 12
Therefore, the answer is: 9 ≤ a < 12

If there are only three positive integer solutions for inequality 3x-a ≤ 0, then the value range of a __

Transposition, 3x ≤ a,
Then x ≤ a
3,
∵ the inequality has only three positive integer solutions
Then the positive integer solution is: 1, 2, 3
So 3 ≤ a
3<4.
The solution is: 9 ≤ a < 12
So the answer is: 9 ≤ a < 12

The equation x ^ 2 + y ^ 2-6x-4y + 12 = 0 (1) of the circle that has passed through the point (4,4) finds the tangent equation of the circle (2) Find the tangent equation of a circle with equal intercept in the positive direction of the two coordinate axes

There are two tangents. One is x = 4, and the known circle equation can be reduced to (x-3) ^ 2 + (Y-2), and 2 = 1. Then the center of the circle is (3,2). It can be seen from the drawing that one is x = 4, and the other is a straight line y = ax + B, which can be calculated from the known center of the circle
First calculate the straight line passing through the center of the circle with equal intercept, and then translate it up and down to get the answer. There are also two existing straight lines y = - x = B
If there is a circle center (3,2), the straight line y = - x + 5 can be translated

Solve the ternary primary equation x + y + Z = 12, x + 2Y + 5Z = 22, x = 4Y

x+y+z=12.(1)
x+2y+5z=22.(2)
x=4y.(3)
(1)*5-(2):
4x+3y=38.(4)
Substitute (3) x = 4Y into (4):
19y = 38, y = 2, substitute (3), x = 8
Substitute x, y into (1): z = 12-8-2 = 2
So x = 8, y = 2, z = 2

The following equations 3x-2y = - 1 and 3x-4y = - 5 are solved by the addition, subtraction and elimination method And 6x + 5Z = 5, 3x + 4Z = - 5

[1] - [2], 2Y = - 1 + 5 = 4
So y = 2
[1] * 2 - [2], 3x = - 2 + 5 = 3
So x = 1
So the solution of the equations is: x = 1, y = 2
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