[MOVED] Absolute-Value Equation/Inequality

I have to solve this problem:

. . .|-2x - 1| > 7

This is what I came up with:

. . .-2x - 1 > 7. . .or. . .-2x - 1 < -7

. . .-2x > 8. . .or. . .-2x < -6

. . .x > -4. . .or. . .x < 3

. . .{x | x > -4. . .or. . .x < 3}

...or (-4, 3)

Is this correct? If not what am I doing wrong please. Thank you.

Think about the basic definition of "greater than" and absolute values. For something to have an absolute value of greater than some number, the insides of the absolute value have to be further from zero than that number. So, for instance:

. . . . .|x| > 3

...means that, whatever x is, |x| must be more than 3, or that |x| must be more than three units from zero. So positive values of x must be greater than 3 (further to the right than three units to the right) and negative values must be less than -3 (further to the left than three units to the left). So the absoluve-value inequality must be split as:

. . . . .x < -3 or x > 3

You appear to have mixed them up as:

. . . . .x > -3 and x < 3

This would be the case for |x| < 3, but not for a "greater than" inequality.

Note: The inequality "x > -4" means "[-4, +infinity)", since x is greater than -4. And "x < 3" means "(-infinity, 3)", since x is less than 3. So your set-notation solution is backwards from your inequality-notation solution.

Eliz.

I see, I drew a number line and I think I got it. Then it should be- infinity,3)U(-4, infinity)

Am I correct?

No: the error came in the very first step. Reversing things at the end doesn't "fix" that.

Also, your new solution means "(-infinity, +infinity)", since 3 is greater than -4. Unless you mean the solution to be "all x", this cannot be correct.

Please review the previous explanation, and start again from the beginning.

Eliz.

silverdragon316 said:

Am I correct?

Click to expand...

Please quit guessing.

|-2x-1| >= 7

Thorough Solution

Understand |x| = x IF x >= 0 Example |4| = 4
Understand |x| = -x IF x < 0 Example |-3| = -(-3) = 3

Solve -2x-1 >= 0

-2x >= 1
x <= -1/2

So,

For x <= -1/2, Solve -2x - 1 >= 7
-2x >= 8
x <= -4
This is applicable ONLY where x <= -1/2

For x > -1/2, Solve -(-2x-1) >= 7
2x+1 >= 7
2x >= 6
x >= 3
This is applicable ONLY where x > -1/2

One more step, writing the solution. Go for it.

silverdragon316 said:

I have to solve this problem:

. . .|-2x - 1| > 7

This is what I came up with:

. . .-2x - 1 > 7. . .or. . .-2x - 1 < -7

. . .-2x > 8. . .or. . .-2x < -6

. . .x > -4. . .or. . .x < 3

. . .{x | x > -4. . .or. . .x < 3}

...or (-4, 3)

Is this correct? If not what am I doing wrong please. Thank you.

Click to expand...

ISs this correct?

Get up to speed, here, silver. We already answered that question TWICE, just not with a simple "yes" or "no". Anyway, would "no" be of any value to you? First off, why would we bother to provide lengthy explanations if your answer was right in the first place? Now it is your turn. Think about it and then you tell us what the right answer is.

Part of the study of mathematics is learning how to organize your mind and to think things through logically. This is what we are trying to pursuade you to do. Please proceed.

silverdragon316 said:

ISs this correct?

Click to expand...

Have you checked your answer at all, such as by graphing or trying points?

(By the way, you might want to review how to solve linear inequalities before proceeding any further. The errors you are making indicate that, if you are reading our replies, you are sufficiently lost as to not recognize what you've been given.)

Note: It is generally better to reply with corrections, as stealth-edits (going back and changing the initial post) tend only to confuse the conversation (as the replies now no longer make any sense).

Thank you for your consideration.

Eliz.

. . .|-2x - 1| > 7

This is what I came up with:

. . .-2x - 1 > 7. . .or. . .-2x - 1 < -7

. . .-2x > 8. . .or. . .-2x < -6

. . .x < -4. . .or. . .x > 3

. . .{x | x < -4. . .or. . .x > 3}

...or (- infinity,-4)U(3, infinity)

When you divide by a negative you have to reverse the signs right?
I have done the problem from scratch again.

Sorry I fustrated you guys but I have been out of school a long time and was never very good at it but I am trying. The concepts take me alot longer. I really only understand by similar examples and repetion. But thank you anyway.

silverdragon316 said:

I have been out of school a long time and was never very good at it but I am trying. The concepts take me alot longer....

Click to expand...

That's perfectly understandable and okay.

But when you're given hints, suggestions, explanations, worked examples, and/or solutions, it would probably be helpful to review those, rather than simply replying with another guess.

You are being provided with these helps to help! :wink:

Eliz.

silverdragon316 said:

I have to solve this problem:

. . .|-2x - 1| > 7

This is what I came up with:

. . .-2x - 1 > 7. . .or. . .-2x - 1 < -7

. . .-2x > 8. . .or. . .-2x < -6

. . .x > -4. . .or. . .x < 3

. . .{x | x > -4. . .or. . .x < 3}

...or (-4, 3)

Is this correct? If not what am I doing wrong please. Thank you.

Click to expand...

I actually do try to listen, I even go to thee web link and print the selections and take them with me to work to read and where ever I go also go over my notes. Thank you very much for your help.

silverdragon316 said:

. . .|-2x - 1| > 7

This is what I came up with:

. . .-2x - 1 > 7. . .or. . .-2x - 1 < -7

. . .-2x > 8. . .or. . .-2x < -6

. . .x < -4. . .or. . .x > 3

. . .{x | x < -4. . .or. . .x > 3}

...or (- infinity,-4)U(3, infinity)

When you divide by a negative you have to reverse the signs right?
I have done the problem from scratch again.

Sorry I fustrated you guys but I have been out of school a long time and was never very good at it but I am trying. The concepts take me alot longer. I really only understand by similar examples and repetion. But thank you anyway.

Click to expand...

Sorry for sounding like a broken record but is this right?

One more thing, then you tell us if you have it.

Try values in the regions so defined.

Try x = 0 -- This is in neither region. It shouldn't work.
Try x = 10 -- Does it work?
Try x = -5 -- Does it work?

Prove to yourself that you have a clue. I think you do. Now convince you.

tkhunny said:

One more thing, then you tell us if you have it.

Try values in the regions so defined.

Try x = 0 -- This is in neither region. It shouldn't work.
Try x = 10 -- Does it work?
Try x = -5 -- Does it work?

Prove to yourself that you have a clue. I think you do. Now convince you.

Click to expand...

I don't seem to understand. Am I suppose to replace x with 5, then 10?

That's what substitution is all about. Didn't anyone ever teach you to check you work? Substitute the values for the variable and see if they work.

The idea is this.
1) We KNOW nothing is happening except at x = 3 and x = -4.
2) This defines three regions of the Real Numbers.
3) If ANYthing behaves in the some way in one of the three regions so defined, then it is reasonable to assume that EVERYthing works the same in that one of the three regions so defined.

So, try x = 0 and determin the behavior of EVERYthing -4 < x < 3

You'll have to decide just exactly what to do with x = 3 and x = -4. with everything else, we can make sweeping conclusions with very little effort.