Question

Write a solution in Interval Notation - (you don't have to help me on all, 1 or 2 is fine c: )

1) | m | -2 > 0

2) | x - 4 | - 3 > 5

3) | 6 + 9x | ≤ 24

4) | 1 - 5a | > 29

Answer 1

QUESTION 1

The given inequality is

$|m|-2>0$

We group like terms to get,

$|m|>2$

This implies that,

$-m>2$ or $m>2$.

We simplify the inequality to get,

$m<-2$ or $m>2$.

We can write this interval notation to get,

$(-\infty,-2)\cup (2,+\infty)$.

QUESTION 2

$|x-4|-3\:>\:5$.

We group like terms to get,

$|x-4|\:>\:5+3$.

$|x-4|\:>\:8$

We split the absolute value sign to get,

$-(x-4)\:>\:8$ or $x-4\:>\:8$

This implies that,

$x-4\:<\:-8$ or $x-4\:>\:8$

$x\:<\:-8+4$ or $x\:>\:8+4$

$x\:<\:-4$ or $x\:>\:12$

We can write this interval notation to get,

$(-\infty,-4)\cup (12,+\infty)$.

QUESTION 3

The given inequality is

$|6+9x|\leq 24$

We split the absolute value sign to obtain,

$-(6+9x)\leq 24$ or $(6+9x)\leq 24$

This simplifies to

$6+9x\ge -24$ and $6+9x\leq 24$

$9x\ge -24-6$ and $9x\leq 24-6$

$9x\ge -30$ and $9x\leq 18$

$x\ge -\frac{10}{3}$ and $x\leq 2$

$-\frac{10}{3}\leq x\leq2$

We write this in interval form  to get,

$[-\frac{10}{3},2]$

QUESTION 4

The given inequality is

$|1-5a|>29$

We split the absolute value sign to get,

$-(1-5a)>29$ or $1-5a>29$

This simplifies to,

$1-5a\:<\:-29$ or $1-5a\:>\:29$

This implies that,

$-5a\:<\:-29-1$ or $-5a\:>\:29-1$

$-5a\:<\:-30$ or $-5a\:>\:28$

$a\:>\:6$ or $a\:<\:-\frac{28}{5}$

We write this in interval notation to get,

$(-\infty,-\frac{28}{5})\cup (6,+\infty)$