Question

|x+2|<=1 |x-3|<2 |x+1|=>3 |x-0,3|<4 |1,7+x|>5 |x+4,8|<=6

Answer 1

1) Ix+2I≤1

-1≤x+2≤1

-1-2≤x≤1-2

-3≤x≤-1

x∈[-3; -1]

2)

Ix-3I<2

-2<x-3<2

-2+3<x<2+3

1<x<5

x∈(1; 5)

3)

Ix+1I≥3

-3≥x+1≥3

-4≥x≥2

x∈(-∞; -4] U [2; ∞)

4)

Ix-0,3I<4

-4+0,3<x<4+0,3

-3,7<x<4,3

x∈(-3,7; 4,3)

5) Ix+1,7I>5

-5-1,7>x>5-1,7

x∈(-∞; -6,7)U(3,3; ∞)

6) Ix+4,8I≤6

-6-4,8≤x≤6-4,8

-10,8≤x≤1,2

x∈[-10,8; 1,2]

Answer 2

To solve these inequalities, we'll break them down one by one:

1. |x + 2| <= 1: The absolute value of any number is always non-negative, so the absolute value of x + 2 can be at most 1. This means -1 <= x + 2 <= 1. Solving for x, we have -3 <= x <= -1.

2. |x - 3| < 2: Similar to the previous case, we have -2 < x - 3 < 2. Adding 3 to all parts of the inequality gives 1 < x < 5.

3. |x + 1| >= 3: The absolute value of x + 1 can be greater than or equal to 3. This leads to two cases: x + 1 >= 3 and x + 1 <= -3. Solving these separately, we get x >= 2 and x <= -4.

4. |x - 0.3| < 4: Following the same process, we get -4 < x - 0.3 < 4. Adding 0.3 to all parts yields -3.7 < x < 4.3.

5. |1.7 + x| > 5: Here, we have two cases: 1.7 + x > 5 and 1.7 + x < -5. Solving these separately, we get x > 3.3 and x < -6.7.

6. |x + 4.8| <= 6: We can rewrite this inequality as -6 <= x + 4.8 <= 6. Subtracting 4.8 from all parts gives -10.8 <= x <= 1.2.

Combining the solutions from each inequality, we can find the overall solution range for x:

-10.8 <= x <= -6.7 or -4 <= x <= -1 or 1.2 <= x <= 4.3.