Check out our College Algebra Course: greenemath.com/College_Algebra.html

I can't overstate how helpful this video was. I struggled to understand this concept until I watched this. Thank you so much!

@Greenemath

10 ай бұрын

You're very welcome!

|x+3| > |2x-1| 4:25 |x+5| + |x-3| > 14 14:11 |3/(x-1)| > 5 (21:33)

@Greenemath

2 жыл бұрын

Thanks for posting.

I watched your other videos on modulus too which made my strengthening of the concepts of these type of questions very easy :) Thank You

@Greenemath

3 жыл бұрын

I'm glad you found the video helpful!

Thanks! This video helped me understand this concept. I have been struggling with this for a while.

@Greenemath

Жыл бұрын

Awesome!

you actually saved my homework thank you so much

@Greenemath

2 жыл бұрын

Awesome!

Quick question, how do we know when the critical points ranges have equal to or just less than / more than, etc. confused about the equal to. Your help will be much appreciated.

@Greenemath

3 ай бұрын

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

I have found this video very helpful, thank you for that. I have one question concerning how to discern between -5 not being included in the first range but included in the middle range and a similar question for +3 in respect of the middle and final columns. I followed everything else, it is just the subtleties of the inclusion/non-inclusion. I would appreciate your help with that please? Thanks again.

@Greenemath

2 жыл бұрын

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

@davegoodo3603

2 жыл бұрын

14 mins and 20 seconds.

@Greenemath

2 жыл бұрын

@@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! 😎

What do you complicate so much:|x+3| > |2x-1| use definition of absolute value and get : - (2x-1 )> (x+3) > (2x-1) solve the two inequalities and then you get from left side x

@Greenemath

2 жыл бұрын

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.

@iBen-ry6pj

9 ай бұрын

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

Why didn't you include -2/3 and 4 in the final range as the solution of the first inequality? Both of them are WITHIN the range. I mean it should be [-2/3, 4]. Please let me know. Thanks in advance.

@Greenemath

4 жыл бұрын

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.

In the first example, please explain how did -3 in the first interval not included?

@Greenemath

2 жыл бұрын

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.

The question applies also for absolute value equations.. , thanks

@Greenemath

3 жыл бұрын

See below

Can u please explain to details about how I will rewrite the absolute of the equations

@Greenemath

3 жыл бұрын

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

Why for the last question, cannot combine the range to make 2/5 to 8/5 but 2/5 to 1, or 1 to 8/5

@Greenemath

3 жыл бұрын

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.

Hi , how can I know - in the table ranges if a number is included or not ?

@Greenemath

3 жыл бұрын

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

@Jggh430

3 жыл бұрын

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?

@Greenemath

3 жыл бұрын

@@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.

I wish our professor took his time to explain this instead of giving us a 3- minute briefing lol

@Greenemath

9 ай бұрын

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

hi @GreeneMath can you do a video on logarithms? very advanced question on logs . I mean VERY VERY ADVANCED LEVEL . please . 3 million indian students preparing for JEE exam would appreciate it

@Greenemath

3 жыл бұрын

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: kzread.info/dash/bejne/doCJlbqddMW2pNI.html

U can use a wavy curve method at 28:06 it is easier than that

@Greenemath

Жыл бұрын

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.

|x+3|>|2x-1| f(x)=|x+3| g(x)=|2x-1| The graphs of the two functions meet at two points, say , A and B A x+3=-2x+1 3x=-2 x=-2/3 B x+3=2x-1 x=4 Answer -2/3

@Greenemath

2 жыл бұрын

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

Thank you so much. This helped me a lot with my homework :)

@Greenemath

4 жыл бұрын

Glad it helped!

It's really helpful.... Your video helps me to clear my confusion 🙂 thank you 😊

@Greenemath

2 жыл бұрын

Glad to hear that!

Is there something on double absolute value inequalities?

@Greenemath

2 жыл бұрын

Yes, here is a video on that: kzread.info/dash/bejne/ZnqL1tCwl5TfY9Y.html

@lanomusambazi8654

2 жыл бұрын

@@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.

@Greenemath

2 жыл бұрын

@@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: kzread.info/dash/bejne/q2F10LKLfbTahqg.html

thank you for the explanation sir, very helpful😍

@Greenemath

3 жыл бұрын

Most welcome!

The fact that you do all kinds

@Greenemath

2 жыл бұрын

Good luck with class.

Awesome

@Greenemath

4 жыл бұрын

Thanks :)

Is this lesson for highschool students?

@Greenemath

2 жыл бұрын

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

yes i'm in the middle of a math exam🙄

@Greenemath

2 жыл бұрын

Well I hope you passed.

If you dont mind, I ask you another question which would be my last one with respect to absolute value: I have seen some people that when they solve equation with absolute value they don't care to CONSIDER the scope in which x can be accepted BEFORE solving it and finding the value(s) x can accept to make the equation true. Let me give you an example: Suppose we have an equation |x+3| = -4x - 22. So, these people start solving it given the fact that |expression| = T -> expression = +-T. So, A) x+3 = -4x - 22 -> 5x = -25 -> x= -5. B) -(x+3) = -4x - 22 -> 3x= -18 -> x= -6 However, I start solving the equation by immediately CONSIDERING the scope in which x can be accepted BEFORE finding x value: A) x+3 >=0 -> x>=-3. Now solving the equation: x+3 = -4x -22 -> 5x= -25 -> x=-5. This value can NOT be accepted because it's out of scope. B) x+3 x x =-6. This value is accepted because it's within the scope. I think what those people do is prone to error. Because we can find answers that do not make the equation true. I wanted to have your opinion about it. Thanks in advance

@Greenemath

4 жыл бұрын

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 :)

@theesunnlightt2268

4 жыл бұрын

@@Greenemath I understand. Sorry about that!

@Greenemath

4 жыл бұрын

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 -(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

@theesunnlightt2268

4 жыл бұрын

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

This method does not work for all of them

@ckdjcj3091

9 ай бұрын

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

@Greenemath

9 ай бұрын

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

Quick question, does a > ∣b∣ imply that a+b≥0?

@Greenemath

6 ай бұрын

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.