Solve the following inequality | 3x | > 1; | X of 2|

Solve the following inequality | 3x | > 1; | X of 2|

From | 3x | 1 to | x | 1 / 3; from | X / 2|
2Y square + 4Y = y + 2, how to calculate this
The formula is first transformed into 2Y + 3y-2 = 0, then (2y-1) (y + 2) = 0, so that Y1 = 1 / 2, y2 = - 2
Solving inequality 2 / 2 x + 5 ≤ (2 / 2 3x + 2) + 1
2 / 2 x + 5 ≤ (2 / 2 3x + 2) + 1
Multiply both sides by two to get
x+10=6
x>=3
Solution equation (2Y + 1) square = 9 (Y-1) square
(2Y + 1) square = 9 (Y-1) square
(2Y + 1) square-9 (Y-1) square = 0
[(2y+1)+3(y-1)][(2y+1)-3(y-1)]=0
(5y-1)(-y+5)=0
∴y=1/5 y=5
x=4huo2/5
There is no real solution to the second order equation!
Four or two fifths
Y = 4 or y = 2 / 5 (two fifths)
(2y+1)²=9(y-1)²
(2y+1)²-9(y-1)²=0
【(2y+1)+3(y-1)】【(2y+1)-3(y-1)】=0
（2y+1+3y-3）(2y+1-3y+3)=0
（5y-2）(-y+4)=0
5y-2 = 0 or - y + 4 = 0
Y = 2 / 5 or y = 4
It is known that the solution set of the linear inequality 3x + A / - 13 ＜ 3-x / 2 with respect to X is x ＜ 7, and the value of a is obtained
Solution! + detailed process + explanation. What's the secret of this topic? Thank you! Although others have asked questions on Baidu, they can't react~
Taking a as a known number, the solution set of inequality is obtained,
And X
Solve equation 1. (X-2) square-3 = 0.2. (x + 1) square = 12
1.（x-2）^2-3=0
（x-2）^2=3
X-2 = positive and negative root sign 3
X1 = 2 + radical 3
X2 = 2-radical 3
2.（X+1)^2=12
X + 1 = plus or minus 2 * root sign 3
X1 = 2 * radical 3-1
X2 = - 2 * radical 3-1
Solving absolute value inequality: 1 < absolute value 3x 4 ≤ 6
I looked for a topic, which should be "3x + 4". This kind of problem needs to be discussed
When 3x + 4 ≤ 0, i.e. x ≤ - 4 / 3
One
The process of solving the equation of 5 (3-2y) - 12 (5-2y) = - 17
Step by step. I don't understand
5(3-2y)-12(5-2y)
=15-10y-60+24y
=-45+14y=-17
14y=45-17=28
y=28/14=2
Y=2
15-10y-60+24y=-17
14y-45=-17
14y=-17+45
14y=28
Y=2
The solution set of inequality | 3x-4 | ≤ 4 is______ .
From the inequality | 3x-4 | ≤ 4, we can get - 4 ≤ 3x-4 ≤ 4 by first removing the absolute value; after the term is shifted, we can get: 0 ≤ 3x ≤ 8, and the solution is: 0 ≤ x ≤ 83. So the answer is: {x | 0 ≤ x ≤ 83}
Solve the equation (X-2) square + y square = 4, x = 2Y + 6
X squared - 2x + 2 + y = 4, so much trouble
If the square of X - 2x + 4 is equal to 4, then the square of X - 2x = 0 and the square of X is nonnegative