Hi, Quddoos,

For this problem, you would split the inequality up into two inequalities, or a compound inequality. When the inequality points toward the absolute value like this, the formula is:

|ax+b| < c becomes

**-c < ax+b < c**

So, for us, c = 12, a = 4, and b = 8. So, we have:

**-12 < 4x+8 < 12**

Then, we solve this for x:

**-12 < 4x+8 < 12**

-8 -8 -8

**-20 < 4x < 4**

/4 /4 /4

**-5 < x < 1**

So, the answer is -5 < x < 1.

Good luck, quddoos, I hope this assists.

Till Then,

Steve

Solve: `|4m+8| < 12`

This means that 4m + 8 is less than 12 units from 0 on a number line, therefore:

`4m + 8 < 12and 4m + 8 > -12`

Solve each.

`4m + 8 < 12` Subtract 8.

`4m < 4` Divide by 4.

`:. m < 1`

`4m + 8 > -12` Subtract 8.

`4m > -20` Divide by 4.

`:. m > -5`

**This makes a final answer of: `-5 < m < 1` **

The absolute value of (4m + 8) is less than 12.

This gives:

-12 < (4m + 8) < 12

-12 < (4m + 8)

= -20 < 4m

= -5 < m

(4m + 8) < 12

= 4m < 4

= m < 1

Therefore the values of m lie between -5 and 1. The solution set of |4m+8|<12 is (-5, 1)

|4m + 8| < 12

Change the equation to

4m + 8 < 12 and 4m + 8 > -12 now subtract 8 to both sides of both equation

By subtracting , you should get

4m < 4 and 4m > -20 now divide by 4 on both sides of both equation .

By dividing , you should get

m < 1 and m > -5

So your answer is

-5 < m < 1

`4m+8lt12 `

`4m+8gt-12 `

`4m+8-8lt12-8 `

`4mlt4 `

`(4m)/4lt4/4 `

`mlt1 `

`4m+8gt-12 `

`4m+8-8gt-12-8 `

`4mgt-20 `

`(4m)/4gt(-20)/4`

`mgt-5 `

`-5ltmlt1 `