SOLUTION: |3x-6| >9 Solve each absolute value inequality. Graph the solution. The bigger sign actually has a line immediately under it. First i added 6 to -6 and to 9 after I got

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Question 1196404: |3x-6| >9 Solve each absolute value inequality. Graph the solution.
The bigger sign actually has a line immediately under it.
First i added 6 to -6 and to 9 after I got 15 from 9+6 I didn’t know what to do I need help understanding the process of solving the question. Please and thank you.
Found 3 solutions by greenestamps, ikleyn, math_tutor2020:
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!

Your inequality is of the form

If you think of as meaning how far A is from zero, the solution set is going to be two distinct parts: either A is positive and greater than or equal to B, or A is negative and equal to or less than ("more negative than") B.

So to find the solution set for this inequality, you need to solve two inequalities:

(1)

and

(2)

Then, since the solution set is two distinct parts, your answer will be in the form "..." OR "...".

It appears from the work you show that you know how to solve each of those inequalities, so I'll let you finish.

Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website!
.

Hello, from your post, I see that you are unfamiliar with the conception of absolute value,
as well as are unfamiliar with solving absolute value equations.

In this site, there is a group of lessons that teach students to this subject.

The introductory lesson is
    - Absolute Value equations

Look in it :   you will find there many solved problems with detailed explanations.

At the end of that lesson, you will find the links to the lessons on solving more complicated absolute value equations.

Also, for those who start learning absolute value equations, it is useful to know how the graphs/plots
look like of absolute value functions.

To learn about it, look into the lessons
    How to plot functions containing Linear Terms under the Absolute Value sign. Lesson 1
    How to plot functions containing Linear Terms under the Absolute Value sign. Lesson 2
    How to plot functions containing Linear Terms under the Absolute Value sign. Lesson 3
in this site.

When you will need more complicated equations/plots, come again and do not hesitate to ask.

Happy learning ( ! )

Answer by math_tutor2020(3816)   (Show Source):
You can put this solution on YOUR website!

Rule to write down in your notebook:
If |A| ≥ B, then either A ≥ B or A ≤ -B

Example:
|x| ≥ 12 means x ≥ 12 or x ≤ -12
|x| represents the distance from x to 0 on the number line. We want the distance to be 12 or larger. So that places us in the interval x ≥ 12 or x ≤ -12
I strongly recommend drawing out a number line to see what's going on.

Anyways, onto the problem at hand:
|3x-6| ≥ 9
3x-6 ≥ 9 or 3x-6 ≤ -9 ..... use the rule mentioned above
3x ≥ 9+6 or 3x ≤ -9+6
3x ≥ 15 or 3x ≤ -3
x ≥ 15/3 or x ≤ -3/3
x ≥ 5 or x ≤ -1
x ≤ -1 or x ≥ 5

x ≤ -1 turned into interval notation is (-∞, -1]
x ≥ 5 turned into interval notation is [5, ∞)
The entire solution set in interval notation is (-∞, -1] U [5, ∞) where the U symbol is the union operator.

To graph this, we'll have closed filled in circles at -1 and 5
The closed circles signal to the reader " include this endpoint in the shaded solution set".
Then we shade everywhere but the region between the closed endpoints
We shade to the left of -1; and to the right of 5

If you were to pick any value from the shaded region (say x=7), then it makes the original inequality true
|3x-6| ≥ 9
|3*7-6| ≥ 9
|21-6| ≥ 9
|15| ≥ 9
15 ≥ 9
The last inequality is true, so the original inequality is true for x = 7.
Try out other x values from the shaded regions to help partially confirm the answer.
Also, try out values from the non-shaded region to see they make the inequality false.

Side note: The keyword "or" is used (not "and") when joining the intervals x ≤ -1 or x ≥ 5. This is to say x could be in either interval.
We can't say x ≤ -1 and x ≥ 5 since x cannot be in both intervals at once.