Solving Advanced Absolute Value Inequalities 

You're very welcome!

Thanks for posting.

Awesome!

You can use something like desmos and graph, this video is on the algebraic method.

Awesome!

I'm glad you found the video helpful!

You would consider the endpoints using the rules for a strict or non-strict inequality.

I'm not complicating anything, you need a strategy that always works. What happens when you try to solve |x + 3| > |2x - 1| + 5? Your method is going to fall apart, whereas what I showed you will always work.

An easy trick is applying the concept that if two numbers x and y are positive and x

See below

If you give me a time marker of the problem, I'll watch that part again and explain what's going on.

14 mins and 20 seconds.

@@davegoodo3603 When you set up intervals for this type of problem, you have to think about where the expression inside of the absolute value bars is negative and then non-negative, which means 0 or positive. That is why you have those specific intervals. In other words, you need -(x + 5) for the interval where x + 5 is negative (x < -5) and x + 5 for the interval where it is non-negative, which starts where x is -5. Then you have the other guy there which follows the same logic. x - 3 is going to be negative when x is strictly less than 3. When x is 3 or larger then x - 3 is going to be non-negative, meaning 0 or some positive value. Hope this helps, you might want to try graphing this problem on Desmos.com to get a picture of what's going on and then try with another example to reinforce the concept. Good luck with your study of math! 😎

GreeneMath.com has every video I have ever made for free. I probably have 10 videos or so on logs. You can also watch this full video if you want: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-DLU3XdDbPto.html

You definitely can't explain this process in 3 minutes. Unless you were solving basic absolute value inequalities, those are pretty simple.

1 is not a solution to the problem, so it needs to be excluded. In the problem, you have x - 1 in the denominator. If x = 1, you will have 0 as the denominator which is undefined.

You need a method that works for all situations. In the practice test, you will see much harder examples and the general method is what is going to work.

You can re-watch 5:05, where the set up is explained. If x + 3 = 0, then x = -3 is a turning point. You want to find the values where x + 3 < 0, which will occur from negative infinity up to but not including -3. From -3 to infinity, this guy is zero or positive.

Hi, we have a strict inequality, so those numbers are NOT in the solution. You can easily prove this by plugging into the original inequality. |x + 3| > |2x - 1|, so let's just use 4 since it's easy to work with. |4 + 3| > |2 * 4 - 1| -> |7| > |7| -> 7 > 7, which is false. 7 is not greater than 7, 7 is equal to 7. If you plug in -2/3, you will find the same thing. Pay close attention to the sign that is being used. In this case we have a strictly greater than, so the number on the left must be greater than the number on the right.

Rewrite it in what way? What exactly are you trying to do or what problem are you trying to solve?

Glad to hear that!

A number is in the interval if it falls between the endpoints of the interval.

yes , that I understood, I meant the borders of the interval - sometimes you said that a number is included and sometimes you didn’t , for example [1/2,3) - the 1/2 is included but the 3 isn’t , I noticed the in the video of the equations you never included any of the borders - (x,y) - in your examples , is it always true?

@@Jggh430When you have an inequality, you consider the endpoints or turning points separately. If your interval works as a solution, look at the endpoints and see if they work. In other words, just plug into the original inequality. That's all you need to do with any inequality.

Yes, here is a video on that: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-4FWtnwg1y3s.html

@@Greenemath Thank you so much sir. I have already watched this one. Very serious disciple following your footsteps to higher and advanced maths. I actually meant double absolute rational inequalities where both rational inequalities are absolute. Do we have such? I really need it. It's like I saw it somewhere but I can't trace the video.

@@lanomusambazi8654 I didn't realize that I linked to the video you commented on, I thought you were on absolute value equations and needed inequalities. I don't have a video with rational inequalities, but I do have one with rational equations. It's kind of basic, but here it is: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-y-AnPRMQtVE.html

Glad it helped!

Good luck with class.

Hi, there's no way I'm reading all of that. I don't mind quick questions, but I'm not a personal tutor. If you want to ask a quick question in regard to the video posted, I'm happy to answer. Good luck with your studies :)

@@Greenemath I understand. Sorry about that!

No problem, I try to answer as many as I can, but long questions are really hard as I only a lot so much time for these each day. I was able to skim read what you wrote and here's a quick note. The examples you are likely looking at are mostly where the x is only inside of the absolute value bars. So: |x + a| = k This sets up as: x + a = k or x + a = -k So in these cases, you won't have solutions that don't work. You don't need to go through all the extra stuff you are talking about. Now in a more complex example, like yours where there is a variable outside of the absolute value bars or double absolute value bars and a lose number, you need to set up intervals on the number line. I'm not sure where you got that x = -6, but that isn't the right solution, however, your thoughts were in the right place. x > 3 x + 3 = -4x - 22 x = -5 Same as yours, reject this solution, it's not in the given interval x < 3 -(x + 3) = -4x - 22 -x - 3 = -4x - 22 -x - 3 + 3 = -4x - 22 + 3 -x = -4x - 19 -x + 4x = -19 3x = -19 x = -19/3 Lies in the given interval, so we can accept this solution. Not sure how you got x = -6, but that isn't correct. But you had the right idea. Check: |-19/3 + 3| = -4(-19/3) - 22 |-19/3 + 9/3| = 76/3 - 22 |-10/3| = 76/3 - 66/3 10/3 = 10/3 True

@@GreenemathThank you! I really appreciate it. Yes I made a mistake while calculating x.

Most welcome!

It's for anyone who wants to take the time to learn. Thanks for watching.

Thanks :)

Well I hope you passed.

Or I’m making a mistake I’m not sure please help me

The method in this video works 100% of the time.

No, you should break this up into cases and then write a rule for each case. Think about the different possible choices for b, positive, negative, zero.