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 -(10)/(3)` and
`x\leq 2`

`-(10)/(3)\leq x\leq2`

We write this in interval form to get,

`[-(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\:<\:-(28)/(5)`

We write this in interval notation to get,

`(-\infty,-(28)/(5))\cup (6,+\infty)`