Question

1052 Решить 1) |x+3| < 4; 2) |x-1| ⩽ 2; 3) |5+x| < 8; 4) |6 - x| ⩽ 2/3; 5) |4-3x|< 4,7; 6) |8. 1/7+5x|< 1;

Answer 1

Ответ:

Ответ на картинке

на 2 и 3 я отвечала

Answer 2

I can help you solve these inequalities. Here are the solutions:

1) $$|x+3|<4$$ This means that $$-4

2) $$|x-1|\leq 2$$ This means that $$-2\leq x-1\leq 2$$ Adding 1 to both sides, we get $$-1\leq x\leq 3$$ The solution set is $$[-1,3]$$

3) $$|5+x|<8$$ This means that $$-8<5+x<8$$ Subtracting 5 from both sides, we get $$-13

4) $$|6-x|\leq \frac{2}{3}$$ This means that $$-\frac{2}{3}\leq 6-x\leq \frac{2}{3}$$ Subtracting 6 from both sides, we get $$-\frac{20}{3}\leq -x\leq -\frac{16}{3}$$ Multiplying by -1 and reversing the inequalities, we get $$\frac{16}{3}\geq x\geq \frac{20}{3}$$ The solution set is $$[\frac{20}{3},\frac{16}{3}]$$

5) $$|4-3x|<4.7$$ This means that $$-4.7<4-3x<4.7$$ Subtracting 4 from both sides, we get $$-8.7<-3x<0.7$$ Dividing by -3 and reversing the inequalities, we get $$-\frac{7}{30}>x>\frac{29}{10}$$ The solution set is $$(-\infty,\frac{29}{10})\cup(-\frac{7}{30},\infty)$$

6) $$|8+\frac{1}{7}5x|<1$$ This means that $$-1<8+\frac{1}{7}5x<1$$ Subtracting 8 from both sides, we get $$-9<\frac{5}{7}x<-7$$ Multiplying by 7 and dividing by 5, we get $$-\frac{63}{5}

If you need more help with solving inequalities, you can check out these websites: [Mathway](https://www.mathway.com/ru/Algebra), [Quickmath](https://quickmath.com/solve/), or [Microsoft Math Solver](https://mathsolver.microsoft.com/ru/algebra-calculator).