Question

Решите неравенства c модулем |5x+3| -4 ≥ 9 |10-4x| ≤ 2 |3+x|+7<10 |4-x| < 5 |3x-9|+2>7 |3x+2| - 1 ≥ 10

Answer 1

& - знак бесконечности.
1) |5x+3|>=13
5x+3>=13 и 5x+3<= -13
5x>=10
x>=2
5x<=-16
x<= -3,2
(-&; -3,2] [2; +&)

2) 10-4x<=2 и 10-4x>=-2
-4x<=-8
x>=2
-4x>=-12
x<=3
[2; 3]

3) 3+x<3 и 3+x>-3
x<0
x> -6
(-6; 2)

4) 4-x<5 и 4-x>-5
-x<1
x> -1
-x>-9
x<9
(-1; 9)

5) 3x-9>5 и 3x-9<-5
3x>14
x>14/3
3x<4
x<4/3
(-&; 4/3) (14/3; +&)

6) 3x+2>=11 и 3x+2<=-11
3x>=9
x>=3
3x<=-13
x<=-13/3
(-&; -13/3] [3; +&)

Answer 2

Solving the Inequalities with Absolute Value

To solve the given inequalities involving absolute values, we will break them down into separate cases based on the sign of the expressions inside the absolute value bars. Let's solve each inequality step by step.

1. |5x + 3| - 4 ≥ 9 |10 - 4x|

We have two absolute value expressions here. Let's consider two cases:

Case 1: (5x + 3) ≥ 0 and (10 - 4x) ≥ 0 - For (5x + 3) ≥ 0, we have x ≥ -3/5. - For (10 - 4x) ≥ 0, we have x ≤ 10/4 = 5/2.

Case 2: (5x + 3) < 0 and (10 - 4x) < 0 - For (5x + 3) < 0, we have x < -3/5. - For (10 - 4x) < 0, we have x > 5/2.

Combining the solutions from both cases, we have -3/5 ≤ x ≤ 5/2.

2. |3 + x| + 7 < 10 |4 - x|

Similar to the previous inequality, we have two absolute value expressions. Let's consider two cases:

Case 1: (3 + x) ≥ 0 and (4 - x) ≥ 0 - For (3 + x) ≥ 0, we have x ≥ -3. - For (4 - x) ≥ 0, we have x ≤ 4.

Case 2: (3 + x) < 0 and (4 - x) < 0 - For (3 + x) < 0, we have x < -3. - For (4 - x) < 0, we have x > 4.

Combining the solutions from both cases, we have x < -3 or x > 4.

3. |3x - 9| + 2 > 7 |3x + 2|

Again, we have two absolute value expressions. Let's consider two cases:

Case 1: (3x - 9) ≥ 0 and (3x + 2) ≥ 0 - For (3x - 9) ≥ 0, we have x ≥ 3. - For (3x + 2) ≥ 0, we have x ≥ -2/3.

Case 2: (3x - 9) < 0 and (3x + 2) < 0 - For (3x - 9) < 0, we have x < 3. - For (3x + 2) < 0, we have x < -2/3.

Combining the solutions from both cases, we have x < -2/3 or x ≥ 3.

4. |3x + 2| - 1 ≥ 10

We have a single absolute value expression here. Let's consider two cases:

Case 1: (3x + 2) ≥ 0 - For (3x + 2) ≥ 0, we have x ≥ -2/3.

Case 2: (3x + 2) < 0 - For (3x + 2) < 0, we have x < -2/3.

Combining the solutions from both cases, we have x < -2/3 or x ≥ -2/3.

Summary of Solutions

1. |5x + 3| - 4 ≥ 9 |10 - 4x|: -3/5 ≤ x ≤ 5/2 2. |3 + x| + 7 < 10 |4 - x|: x < -3 or x > 4 3. |3x - 9| + 2 > 7 |3x + 2|: x < -2/3 or x ≥ 3 4. |3x + 2| - 1 ≥ 10: x < -2/3 or x ≥ -2/3

Please note that these solutions are based on the given inequalities and the assumptions made during the solution process.