Question

3) |x +1|>3;|х-0,3|<4​

Answer 1

$|x+1|>3\\ \left[\begin{array}{ccc}x+1>3\\ \\ x+12\\ \\ x$

Відповідь: x ∈ (-∞; -4) ∪ (2; +∞).

|x - 0,3| < 4

-4 < x - 0,3 < 4

-3,7 < x < 4,3

Answer 2

To solve the inequalities, we'll consider them one by one:

1. |x + 1| > 3

The absolute value of a number is always non-negative, meaning it is either positive or zero. For the expression |x + 1| to be greater than 3, it means that the value inside the absolute value brackets must be either greater than 3 or less than -3.

So we have two cases to consider:

Case 1: x + 1 > 3 Solving for x: x > 3 - 1 x > 2

Case 2: -(x + 1) > 3 Solving for x: x + 1 < -3 x < -3 - 1 x < -4

Now we have the range of x values for which |x + 1| > 3: x < -4 or x > 2.

2. |x - 0.3| < 4

Similar to the previous inequality, we'll consider two cases:

Case 1: x - 0.3 < 4 Solving for x: x < 4 + 0.3 x < 4.3

Case 2: -(x - 0.3) < 4 Solving for x: x - 0.3 > -4 x > -4 + 0.3 x > -3.7

Now we have the range of x values for which |x - 0.3| < 4: -3.7 < x < 4.3.

Combining the results from both inequalities, we find the intersection of the two solution sets:

-3.7 < x < -4 or x > 2

However, it's important to note that there is no intersection between the two solution sets. Therefore, the final solution is:

-3.7 < x < -4 or x > 2