Question

|5-3x|> 1 |x×2|>_ 0 |3x-7|>_ -8

Answer 1

|5-3x|> 1
5-3x>1 и 5-3х<-1
3х<4. 3х>6
х<1⅓. х>2
х€(-∞;1⅓)v(2;+∞)


|x×2|≥0
x€R
(|f(x)|≥0 ∀x€R)

|3x-7|≥-8
x€R
(|f(x)|≥-8 ∀x€R)

Answer 2

To solve the given inequalities, we'll break it down into three separate inequalities and solve each one step by step:

1. |5 - 3x| > 1
2. |x * 2| ≥ 0 (Note: Any absolute value is always non-negative, so this inequality holds for all real values of x.)
3. |3x - 7| ≥ -8 (Note: Any absolute value is always non-negative, so this inequality holds for all real values of x.)

Let's solve the first inequality:

1. |5 - 3x| > 1

We'll consider two cases:

Case 1: (5 - 3x) > 1 5 - 3x > 1 -3x > 1 - 5 -3x > -4 Divide by -3 (Note: When dividing by a negative number, the inequality sign flips) x < 4/3

Case 2: -(5 - 3x) > 1 -5 + 3x > 1 3x > 1 + 5 3x > 6 Divide by 3 x > 2

So, the solution to the first inequality is 2 < x < 4/3.

Now, let's move on to the third inequality:

3. |3x - 7| ≥ -8

Since any absolute value is non-negative, this inequality holds for all real values of x.

Now, let's consider the second inequality:

2. |x * 2| ≥ 0

As mentioned earlier, any absolute value is always non-negative, so this inequality is true for all real values of x.

In summary, the solution to the given system of inequalities is: 2 < x < 4/3

Note: The second and third inequalities are always true for any real value of x, so they don't restrict the solution set. The only constraint comes from the first inequality.