Question

2 – 3х > 3х + 1 3 – х ≤ 4 + 6х

Answer 1

ОТВЕТ

1.

$x < \frac{1}{ 6}$

2.

$x \geqslant - \frac{1}{7}$

Пошаговое объяснение:

1.

$2 - 3x > 3x + 1$

$- 3x - 3x > 1 - 2$

$- 6x > - 1$

$x < \frac{1}{6}$

2.

$3 - x \leqslant 4 \times 6x$

$- x - 6x \leqslant 4 - 3$

$- 7x \leqslant 1$

$x \geqslant - \frac{1}{7}$

Answer 2

1) 2 - 3x > 3x + 1
3x + 3x < 2 - 1
6x < 1
x < 1/6
2) 3 - x <= 4 + 6x
6x + x => 3 - 4
7x => - 1
x => - 1/7
Ответ: 1) (- бесконечности; 1/6); 2) [- 1/7; + бесконечности).

Answer 3

To solve the inequalities, let's start with the first one:

2 - 3x > 3x + 1

Let's simplify it step by step:

1. Move the variable terms to one side of the inequality and the constant terms to the other side:

2 - 1 > 3x + 3x

1 > 6x

2. Divide both sides by 6, remembering to flip the inequality sign when dividing by a negative number:

1/6 > x

x < 1/6

So the solution to the first inequality is x is less than 1/6.

Now let's move on to the second inequality:

3 - x ≤ 4 + 6x

Again, simplify it step by step:

1. Move the variable terms to one side and the constant terms to the other side:

3 - 4 ≤ 6x + x

-1 ≤ 7x

2. Divide both sides by 7, remembering to flip the inequality sign when dividing by a negative number:

-1/7 ≤ x

x ≥ -1/7

So the solution to the second inequality is x is greater than or equal to -1/7.

To summarize:

• The solution to the first inequality, 2 - 3x > 3x + 1, is x < 1/6.
• The solution to the second inequality, 3 - x ≤ 4 + 6x, is x ≥ -1/7.