Question

1)12+х>18 2)6–х≤4 3)1+3х>10 4) 3х+8<0 5)6+х<3–2х 6)4+12х>7+13х 7)3(2+х)>4-х 8)–(4–х)≤2(3+х)

Answer 1

Ответ:

:x=6

x=10

x=3

x=8/3

x=3далле незн

Answer 2

Sure, I can help you solve these inequalities. Let's go through each one step by step:

1. $12 + x > 18$ Subtract 12 from both sides: $x > 6$

2. $6 - x \leq 4$ Subtract 6 from both sides: $-x \leq -2$ Divide by -1 (remember that dividing by a negative number reverses the inequality): $x \geq 2$

3. $1 + 3x > 10$ Subtract 1 from both sides: $3x > 9$ Divide by 3: $x > 3$

4. $3x + 8 < 0$ Subtract 8 from both sides: $3x < -8$ Divide by 3: $x < -\frac{8}{3}$

5. $6 + x < 3 - 2x$ Add $2x$ to both sides and subtract 6 from both sides: $3x < -3$ Divide by 3: $x < -1$

6. $4 + 12x > 7 + 13x$ Subtract $12x$ from both sides and subtract 7 from both sides: $-x > -3$ Multiply by -1 (remember that multiplying by a negative number reverses the inequality): $x < 3$

7. $3(2 + x) > 4 - x$ Distribute the 3 on the left side: $6 + 3x > 4 - x$ Add $x$ to both sides: $4x > -2$ Divide by 4: $x > -\frac{1}{2}$

8. $-(4 - x) \leq 2(3 + x)$ Distribute the -1 on the left side and distribute the 2 on the right side: $-4 + x \leq 6 + 2x$ Subtract $x$ from both sides: $-4 \leq 6 + x$ Subtract 6 from both sides: $-10 \leq x$

So, summarizing the solutions for each inequality:

1. $x > 6$
2. $x \geq 2$
3. $x > 3$
4. $x < -\frac{8}{3}$
5. $x < -1$
6. $x < 3$
7. $x > -\frac{1}{2}$
8. $x \geq -10$