Finally I have understood double absolute value equations from your explanation. Only understanding it today after 2 years of serious struggle. Thank you so much sir. May the Lord remember you.
Imagine I'm doing the calculations right now with great confidence indeed. Thank you once again I'm so happy to know this advanced math of absolute equations. The next one I want is absolute inequalities.
How is someone this smart to know these things they don't teach in class? THANK YOU. Such a great video. never thought a 40 minute math video could pas by so fast
Omg thank you for your existence!!! Our great maths teacher was given a transfer and we now have a terrible temporary maths teacher who literally didn't do a single question on this and expects us to just know it. So thank you for saving my exam ❤️
Really Helpful! I was trying to figure out how to solve absolute value equations with two absolute value operations and a loose number and I found this gem. Thank you!
Omg! I was always terrified by absolute values problems this teacher made me realized that there is nothing to it. I understood the concept and I feel like I can solve any absolute value problem thrown at me. THANK YOU so much you are such a gifted teacher. I wish you could be my math teacher.
Would you please let me know why you didn't include 1/3 and 1 - the critical points- at 11:20 and you wrote (1/3, 1) instead of [1/3, 1) for the middle section and for the last one [1, infinity)?
When one of the absolute value expressions is zero, it's a turning point, going from neg to pos or pos to neg. So when one expression is zero, the expression itself is going to be gone. Think about the scenario where x = 1, |x - 1| + |3x - 1| = 5, so we know that x - 1 will be 0, and 3x - 1 will be positive, so now we have 3x - 1 = 5, which gives us a solution of x = 2, which doesn't work since we explicitly said that x = 1. Also, you could just say does 3(1) - 1 = 5 -> 2 = 5, so the answer is no...
I had never seen these approaches before. I have found that making graphs removes a lot of the mystery. they tell very quickly the number of solutions I should expect.
You can definitely graph each part and then find the points where they intersect. You can also solve the four possible equations and then check to throw out two extraneous solutions. Lots of ways to solve that type of equation.
1) 12:00 so for that left side we are putting negative sign outside because we know what value makes the number negative inside the interval so we are using negative sign outside to make it positive - - = +? 2) What are we doing on middle part 13:00? are we trying to find intervals that make the expression 0? Ok for (3x - 1) we know 1/3 makes it 0 so any other value GREATER is going to be positive. But for (x-1) we know 1 makes it 0 but any value LESS than 1 is negative but why can we not say any value greater than 1 will result in positive like (3x-1)? Why we say GREATER value for that and LESS value for (x - 1) My random guess: (I am guessing the left side of line we focused on value that makes the expression negative which was 1/3 so we now went left to right so we said positive but for right side we are focusing on value that is greater than 1 to keep it positive for 1/3 or 1 so now we going right to left or negative?) 3) Are we picking left intervals that have less value than the other one like in 12:00 and 22:17? and right side we pick bigger interval number like 1/3 is less than 1 so we put 1/3 on left and 1 on right? Thanks Teacher Green
Follow the steps given in the tutorial. Here is a written lesson, it might be easier to follow. greenemath.com/College_Algebra/71/Absolute-Value-Equations-3Lesson.html
So these 22:27 or 11:09 you put negative infinity and smallest number on interval on left? And middle stays the same? and the right part includes the largest number and positive infinity interval?
Follow the steps given in the tutorial. Here is a written lesson, it might be easier to follow. greenemath.com/College_Algebra/71/Absolute-Value-Equations-3Lesson.html
Could you please confirm this, because I think I found an easier method of solving the questions with 2 absolute values and the extra bit but the method makes no sense. So basically I tried to just put the extra bit on one side of the equation and the 2 absolute values on one side and made one equation with positive value of the extra bit and one equation with the negative value of the extra bit, taking the absolute values as a positive value. For each question that i tried it had worked, could you please explain why
Your method isn't going to work. That only works if you have a single absolute value expression. |ax + b| = k, then ax + b = k or ax + b = -k, what you are trying to set up does not work because you have two absolute value expressions and a loose number. You need to either set up four different possible equations and then check for extraneous solutions or use the method proposed here which develops a rule for dropping the absolute value bars and checking to see if the solution lies in the correct interval.
@@Greenemath Thats what I thought too, however when i tried both the 4 different equations and also my method. The method I used gave me the 2 different answers completely disregarding the ones that didn't work. I have confirmed this because I tried it with many different equations and it worked correctly every single time. I could give you an example if you would like, but the method is quite simple so you could try it too.
@aaryan7490 If you want to post some examples, I can show you that your method does not work. You can't set the sum or difference of two absolute value expressions equal to the number and then the negative of the number to solve that type of equation. That absolutely will not work.
@@Greenemath never mind I have confirmed with a friend of mine, and by chance I was solving equations that could be factorized into the same coefficients of x which then in turn the positive value and the negative value of it would cancel out hence the 2 positive value giving me the correct answer. Sorry for the misunderstanding
Hello. I was confused why you don't use closed bracket instead of parentheses. What if we get the zeros of the absolute values as a solution? (Like in the third question when you construct table) Please reply as soon as possible.
You might find this example helpful. ua-cam.com/video/3a5nda4_nSs/v-deo.html You probably need to watch the course in order. I address that exact issue in a previous lesson.
Hi! the video was really helpful, thanks a lot for the tutorial...I just had one doubt in the questions with double mods, if the value of x in any interval coincides with one of the bounds of it do we consider it or not? For eg. at 14:48, in case the value of x comes 1/3 itself, would we include it in our solns??
I would give the video another watch. I spoke about this when setting up the intervals. Write out the problem and think about what happens when x = 1/3? |3 * 1/3 - 1|, is the expression going to be negative, 0, or positive?
Hi i dont know if you are still seeing these comments, but im struggling to get the part where i find x for both equations, how do i know if there is a solution or no? Now that seems to depend on conditions but how do i know which conditions do i need to set?
I'm not exactly sure what you are asking. I would try re-watching the tutorial and working the problems with me as I explain. If you have a specific question about a step in a problem, leave a time marker and I'm happy to answer. Here is a written tutorial if that works better: greenemath.com/College_Algebra/71/Absolute-Value-Equations-3Lesson.html
@@Greenemath Okay, ill try to be more specific, in an example where there is absolute value, before we make 2 equations we are supposed to add some conditions right? Now im wondering how do i set those conditions, based on what?
These two courses will help if you have time to watch them, they cover all the different types of Absolute Value Equations and Inequalities from beginner to advanced: ua-cam.com/video/kBu_HpegOC8/v-deo.htmlsi=RiX9sg29KFrejylW ua-cam.com/video/3a5nda4_nSs/v-deo.htmlsi=SfN-aGPAVDLeOgHj
23:37 so in the lower part of the table, will the order always be : Neg. Pos. Pos Neg. Neg. Pos…? I’m not trying to fully understand, (since that’s a bit beyond my paygrade) but I’m trying to memorize the process since it’s easier
@mannyballon2837 This company makes those but they're really made for artists. It's kind of overdoing it but I really like using it. estore.wacom.com/en-us/ estore.wacom.com/en-us/wacom-cintiq-22-dtk2260k0a.html I have the older version of the one above, again you can just use a tablet at this point. That didn't exist when I started the project.
It's in the video. Basically, you set each expression inside the absolute value bars equal to zero and solve. This will give you the break points. This video is a bit more detailed if you want to watch: ua-cam.com/video/BEIzTKoTM8Y/v-deo.html
we have the same answer but there are times that, when i tried to verify it through computations, it's giving me the wrong answer. Can you please make a video where you verify the answer? I just need it ty
That's going to take quite a while to solve by hand. Simpler way? I don't know of any, I've never seen a problem such as that given for homework. I would use a computer to solve it in real life.
I just saw the size of that and was immediately thinking to use Symbolab or Wolfram to get the answer. I just graphed it in desmos and got x = 0 as well.
@@Greenemath thank you 🙌. I put -10 and 2 to the place of x, and I got 5. Both of them are correct. Actually in my task I was asked to find the sum of roots. So the answer will be -10+2= -8 , right?
@@Greenemath Oh, I think I wrote my question wrongly. I'm not so good at english 😅. In the task it was asked to find the sum of all answers of x. I wrote "root" , translated wrong ))
@@slowcat-j No, your English is perfectly fine. The answer, solution, or root all mean the same thing in Math. Basically, what values can replace the variable and give you a true statement.
Yes, but it would take longer. You would have to think about all the possibilities. If you had something like: |ax + b| + |cx + d| = e Then you would have to do: ax + b + cx + d = e -(ax + b) + cx + d = e -(ax + b) - (cx + d) = e ax + b - (cx + d) = e This will give you extraneous solutions, so you have to go back and check each solution. This type of equation is only going to have two solutions, so two will work and two will have to be thrown out. The method in the video is much quicker.
That part of the problem is purely scratch work. You may include them if you want but be very careful about how you set things up. You can do this: (-infinity, -3/5) | [-3/5, 7/2) | [7/2, infinity) or (-infinity, -3/5) | (-3/5, 7/2) | (7/2, infinity) It is not going to matter at all for an equation as long as your solution is not a breakpoint. When it is, you can just check in the original problem. Here is an example of that scenario from the practice test: ua-cam.com/video/BEIzTKoTM8Y/v-deo.htmlsi=rctTs0VHRJY0nIsJ&t=1516
@@justviewer5458 That's a much easier case, you don't need to follow the thought process from this video. For your scenario you just need two set ups: x - 3 = 2x - 3 x - 3 = -(2x - 3) x = 2, 0 The problems in this video are different, they deal with having a loose number, so the two absolute value expressions can't be set equal to each other.
@@vigneshv3846 Just keep practicing then. Math comes over time. I have a free website GreeneMath.com that has all the videos with practice if you want to check it out. There's also KhanAcademy.org of course and CK-12.org if you like to read.
@@Greenemath Yes, I'm studying linear algebra right now and some matrices are just... Let's just say they are a bit crazy. Loved your video, saw the inequalities one too. Great content, and thank you.
Finally I have understood double absolute value equations from your explanation. Only understanding it today after 2 years of serious struggle. Thank you so much sir. May the Lord remember you.
Glad it helped!
Imagine I'm doing the calculations right now with great confidence indeed. Thank you once again I'm so happy to know this advanced math of absolute equations. The next one I want is absolute inequalities.
Absolute Value Inequalities are solved using a similar strategy, here is the video:
ua-cam.com/video/4FWtnwg1y3s/v-deo.html
@@Greenemath thank you so much for the video. This is indeed great knowledge sir. May you live long.
How is someone this smart to know these things they don't teach in class? THANK YOU. Such a great video. never thought a 40 minute math video could pas by so fast
Glad it was helpful!
Omg thank you for your existence!!! Our great maths teacher was given a transfer and we now have a terrible temporary maths teacher who literally didn't do a single question on this and expects us to just know it. So thank you for saving my exam ❤️
You are so welcome!
Really Helpful! I was trying to figure out how to solve absolute value equations with two absolute value operations and a loose number and I found this gem. Thank you!
You are very welcome, I'm glad you found the video helpful! :)
Omg! I was always terrified by absolute values problems this teacher made me realized that there is nothing to it. I understood the concept and I feel like I can solve any absolute value problem thrown at me. THANK YOU so much you are such a gifted teacher. I wish you could be my math teacher.
You're very welcome!
really really helpful.I'm feeling that you are here to show us the brightest way.May you live long! From Ethiopia
Thanks for the nice comment.
same,from Ethiopia too
Dude this helped so much. You deserve WAY more subscribers. Thanks for the help!
I'm glad it helped you!
Would you please let me know why you didn't include 1/3 and 1 - the critical points- at 11:20 and you wrote (1/3, 1) instead of [1/3, 1) for the middle section and for the last one [1, infinity)?
When one of the absolute value expressions is zero, it's a turning point, going from neg to pos or pos to neg. So when one expression is zero, the expression itself is going to be gone. Think about the scenario where x = 1, |x - 1| + |3x - 1| = 5, so we know that x - 1 will be 0, and 3x - 1 will be positive, so now we have 3x - 1 = 5, which gives us a solution of x = 2, which doesn't work since we explicitly said that x = 1. Also, you could just say does 3(1) - 1 = 5 -> 2 = 5, so the answer is no...
I had never seen these approaches before. I have found that making graphs removes a lot of the mystery. they tell very quickly the number of solutions I should expect.
You can definitely graph each part and then find the points where they intersect. You can also solve the four possible equations and then check to throw out two extraneous solutions. Lots of ways to solve that type of equation.
Thank you so much! I forgot a lot during quarantine and this not only helped me remember it also taught me more
Glad it helped :)
Thanks for the great explanation, helped me clear a lot of confusion I had about absolute value equations
Glad it helped!
after 2 days I finally get it.Tnxxxx
I'm glad to hear that!
What a good video. I like your step-by-step and simplistic approach.
Glad it was helpful!
Is it the same process when dealing with absolute value inequalitys?
And do you have a video specific for that, let me know
ua-cam.com/video/4FWtnwg1y3s/v-deo.html
The video was tremendously helpful! Thank you very much GreeneMath!
Glad it was helpful!
Best video on all of UA-cam.
Doubtful, but thanks for the nice comment.
Seeing sign charts in other than rational inequalities makes me happy :)))
Glad to hear you liked the tutorial!
very detailed, explained well and very helpful, thank you!
Glad it was helpful!
1) 12:00 so for that left side we are putting negative sign outside because we know what value makes the number negative inside the interval so we are using negative sign outside to make it positive - - = +?
2) What are we doing on middle part 13:00? are we trying to find intervals that make the expression 0? Ok for (3x - 1) we know 1/3 makes it 0 so any other value GREATER is going to be positive. But for (x-1) we know 1 makes it 0 but any value LESS than 1 is negative but why can we not say any value greater than 1 will result in positive like (3x-1)? Why we say GREATER value for that and LESS value for (x - 1)
My random guess: (I am guessing the left side of line we focused on value that makes the expression negative which was 1/3 so we now went left to right so we said positive but for right side we are focusing on value that is greater than 1 to keep it positive for 1/3 or 1 so now we going right to left or negative?)
3) Are we picking left intervals that have less value than the other one like in 12:00 and 22:17? and right side we pick bigger interval number like 1/3 is less than 1 so we put 1/3 on left and 1 on right?
Thanks Teacher Green
Follow the steps given in the tutorial. Here is a written lesson, it might be easier to follow.
greenemath.com/College_Algebra/71/Absolute-Value-Equations-3Lesson.html
So these 22:27 or 11:09 you put negative infinity and smallest number on interval on left? And middle stays the same? and the right part includes the largest number and positive infinity interval?
Follow the steps given in the tutorial. Here is a written lesson, it might be easier to follow.
greenemath.com/College_Algebra/71/Absolute-Value-Equations-3Lesson.html
But why did the department of education remove the absolute values from the high school syllabus in South Africa? 😢😢😢
Those problems are rarely taught anywhere. I'm not sure why they were taken out. They only teach the simple type absolute value equations.
thank you for the explanation sir, very helpful.
Glad it was helpful!
but if have got both 2 equations correct we take the 4 equations or what?
I have no idea what you are asking? If you clarify your question, I'm happy to answer.
Here is where l understood double or triple or more absolute value equations clear-crystal
Glad the video was helpful!
thank you, you are really good at explaining
Glad it was helpful! Good luck with your studies :)
Could you please confirm this, because I think I found an easier method of solving the questions with 2 absolute values and the extra bit but the method makes no sense.
So basically I tried to just put the extra bit on one side of the equation and the 2 absolute values on one side and made one equation with positive value of the extra bit and one equation with the negative value of the extra bit, taking the absolute values as a positive value. For each question that i tried it had worked, could you please explain why
Your method isn't going to work. That only works if you have a single absolute value expression. |ax + b| = k, then ax + b = k or ax + b = -k, what you are trying to set up does not work because you have two absolute value expressions and a loose number. You need to either set up four different possible equations and then check for extraneous solutions or use the method proposed here which develops a rule for dropping the absolute value bars and checking to see if the solution lies in the correct interval.
@@Greenemath Thats what I thought too, however when i tried both the 4 different equations and also my method. The method I used gave me the 2 different answers completely disregarding the ones that didn't work. I have confirmed this because I tried it with many different equations and it worked correctly every single time. I could give you an example if you would like, but the method is quite simple so you could try it too.
@aaryan7490 If you want to post some examples, I can show you that your method does not work. You can't set the sum or difference of two absolute value expressions equal to the number and then the negative of the number to solve that type of equation. That absolutely will not work.
@@Greenemath never mind I have confirmed with a friend of mine, and by chance I was solving equations that could be factorized into the same coefficients of x which then in turn the positive value and the negative value of it would cancel out hence the 2 positive value giving me the correct answer. Sorry for the misunderstanding
@aaryan7490 It's cool, best of luck with your studies!
It was really helpful new subscriber from Algeria
Awesome! Thank you!
May someone please share the link to the previous lessons video on this section
ua-cam.com/video/kBu_HpegOC8/v-deo.html
ua-cam.com/video/3a5nda4_nSs/v-deo.html
Ty man saved my grade keep puttin in the work
You are welcome.
Hello. I was confused why you don't use closed bracket instead of parentheses. What if we get the zeros of the absolute values as a solution? (Like in the third question when you construct table)
Please reply as soon as possible.
You might find this example helpful.
ua-cam.com/video/3a5nda4_nSs/v-deo.html
You probably need to watch the course in order. I address that exact issue in a previous lesson.
Thank you that's was very clear and easy to me to understand.
Glad it was helpful!
Hi! the video was really helpful, thanks a lot for the tutorial...I just had one doubt in the questions with double mods, if the value of x in any interval coincides with one of the bounds of it do we consider it or not? For eg. at 14:48, in case the value of x comes 1/3 itself, would we include it in our solns??
I would give the video another watch. I spoke about this when setting up the intervals. Write out the problem and think about what happens when x = 1/3? |3 * 1/3 - 1|, is the expression going to be negative, 0, or positive?
Thank u very much it helps me to final exam
Glad to hear that and good luck :)
Hi i dont know if you are still seeing these comments, but im struggling to get the part where i find x for both equations, how do i know if there is a solution or no? Now that seems to depend on conditions but how do i know which conditions do i need to set?
I'm not exactly sure what you are asking. I would try re-watching the tutorial and working the problems with me as I explain. If you have a specific question about a step in a problem, leave a time marker and I'm happy to answer. Here is a written tutorial if that works better:
greenemath.com/College_Algebra/71/Absolute-Value-Equations-3Lesson.html
@@Greenemath Okay, ill try to be more specific, in an example where there is absolute value, before we make 2 equations we are supposed to add some conditions right? Now im wondering how do i set those conditions, based on what?
These two courses will help if you have time to watch them, they cover all the different types of Absolute Value Equations and Inequalities from beginner to advanced:
ua-cam.com/video/kBu_HpegOC8/v-deo.htmlsi=RiX9sg29KFrejylW
ua-cam.com/video/3a5nda4_nSs/v-deo.htmlsi=SfN-aGPAVDLeOgHj
@@Greenemath Alright thank you
@vesc23 You are very welcome, if you watch the course and have a specific question, leave a time marker and I'll try to answer.
Thank you so much, it was really helpful for me!!
Glad it helped!
A little to fast but excellent keep up the good work greetings from mexico
Thanks! You can always slow the player down.
Great work, very interesting; thanks for sharing.
Glad you enjoyed it!
Great video! Helped a lot!
Glad it helped!
23:37 so in the lower part of the table, will the order always be :
Neg. Pos. Pos
Neg. Neg. Pos…?
I’m not trying to fully understand, (since that’s a bit beyond my paygrade) but I’m trying to memorize the process since it’s easier
There's not really anything to memorize. Just set up the intervals and do the test.
What digital board did you use for this video?
I think you are referring to what I am writing on? That is a Wacom tablet but you can use any tablet you want as long as it has a pencil.
@@Greenemath I want a digital board for PC.
@mannyballon2837 This company makes those but they're really made for artists. It's kind of overdoing it but I really like using it.
estore.wacom.com/en-us/
estore.wacom.com/en-us/wacom-cintiq-22-dtk2260k0a.html
I have the older version of the one above, again you can just use a tablet at this point. That didn't exist when I started the project.
Is this for Pre-Algebra or Algebra 1?
Absolute value equations are normally taught in any algebra/pre-calc course. I've never seen it in pre-algebra, but every class is different.
Very well explained
Glad it was helpful!
Thanks for the explanation
You're welcome!
where are you getting those intervals?
It's in the video. Basically, you set each expression inside the absolute value bars equal to zero and solve. This will give you the break points.
This video is a bit more detailed if you want to watch:
ua-cam.com/video/BEIzTKoTM8Y/v-deo.html
How to plot the values of x in a number line?
For an equation? You just draw a horizontal number line, find your solutions and fill them in.
we have the same answer but there are times that, when i tried to verify it through computations, it's giving me the wrong answer. Can you please make a video where you verify the answer? I just need it ty
Just plug your solution in for x and see if the left and right sides are equal.
Thank you so much I truly appreciate you
You are so welcome
Really helpful . Like it
Glad it was helpful!
|x-3|+|x-2|+|x-1|+|x+1|+|x+2|+|x+3|+|x|=12. I'm using a similar strategy as in 11:20 but I'm wondering if there is a simpler way?
That's going to take quite a while to solve by hand. Simpler way? I don't know of any, I've never seen a problem such as that given for homework. I would use a computer to solve it in real life.
@@Greenemath Hello, thanks for your response! So I figured out the problem. It's a trick question kind of. There is only one solution, x=0.
I just saw the size of that and was immediately thinking to use Symbolab or Wolfram to get the answer. I just graphed it in desmos and got x = 0 as well.
this is really good! tysm!!
I'm glad you like it
Really helpful!
Glad it was helpful!
wat if the x cancels out?
What's the problem?
great help. thank you!
You're welcome!
Thanks!
No problem!
thanks!
You are welcome!
can you help to solve this one |х-2|+|х+4| - |х-3| = 5 ???
Basically, you want to follow the same steps and set up intervals. The answer is going to be x = -10 or x = 2.
@@Greenemath thank you 🙌. I put -10 and 2 to the place of x, and I got 5. Both of them are correct. Actually in my task I was asked to find the sum of roots. So the answer will be -10+2= -8 , right?
@@slowcat-j Yes, but they usually ask that question with quadratics and higher. There is a specific formula you can use on those guys.
@@Greenemath Oh, I think I wrote my question wrongly. I'm not so good at english 😅. In the task it was asked to find the sum of all answers of x. I wrote "root" , translated wrong ))
@@slowcat-j No, your English is perfectly fine. The answer, solution, or root all mean the same thing in Math. Basically, what values can replace the variable and give you a true statement.
what????????/ confuse method is there any way solving in case method
Yes, but it would take longer. You would have to think about all the possibilities. If you had something like:
|ax + b| + |cx + d| = e
Then you would have to do:
ax + b + cx + d = e
-(ax + b) + cx + d = e
-(ax + b) - (cx + d) = e
ax + b - (cx + d) = e
This will give you extraneous solutions, so you have to go back and check each solution. This type of equation is only going to have two solutions, so two will work and two will have to be thrown out. The method in the video is much quicker.
22:24 why neither included? shouldn't they be included somewhere.
That part of the problem is purely scratch work. You may include them if you want but be very careful about how you set things up. You can do this:
(-infinity, -3/5) | [-3/5, 7/2) | [7/2, infinity)
or
(-infinity, -3/5) | (-3/5, 7/2) | (7/2, infinity)
It is not going to matter at all for an equation as long as your solution is not a breakpoint. When it is, you can just check in the original problem.
Here is an example of that scenario from the practice test:
ua-cam.com/video/BEIzTKoTM8Y/v-deo.htmlsi=rctTs0VHRJY0nIsJ&t=1516
@@Greenemath thanks for the reply. i get it now
@fenfox I'm really glad to hear that!
i was waiting to get stick bugged
No idea what that means, but thanks for watching.
@@Greenemath Hahah 😅 its a meme
But the video was very helpful XD
@@coachoikawa4951 Oh I see, well thanks for watching :)
gr8 video
Glad you enjoyed it
9:00
0:00
Easy techniques ,
can i also instead of using + then - , can i just square both sides if the case is like this
|x-3| = |2x-3|
?
@@justviewer5458 That's a much easier case, you don't need to follow the thought process from this video. For your scenario you just need two set ups:
x - 3 = 2x - 3
x - 3 = -(2x - 3)
x = 2, 0
The problems in this video are different, they deal with having a loose number, so the two absolute value expressions can't be set equal to each other.
37:52 I can change any quadratic absolute value equation to "u" then solve?
You can always make valid substitutions but it's not something you would do in every case.
👌
Thanks for watching :)
sorry its correct sorry for misconception
It's okay, as long as you are learning. Good luck with your studies :)
Need more subs
It's a math channel, the subs don't really watch new videos anyway. Good luck with your studies :)
lol your two looks like a derivative symbol. nice lecture though
Yes, like every other mathematician I have terrible handwriting.
@@Greenemath Then I too wish to have a terrible handwriting one day!!
@@vigneshv3846 Just keep practicing then. Math comes over time. I have a free website GreeneMath.com that has all the videos with practice if you want to check it out. There's also KhanAcademy.org of course and CK-12.org if you like to read.
Fellow student here!
Very annoying topic indeed.
Yeah, these are very tedious problems. Unfortunately as you progress in math you find these more often. Good luck!
@@Greenemath Yes, I'm studying linear algebra right now and some matrices are just... Let's just say they are a bit crazy. Loved your video, saw the inequalities one too. Great content, and thank you.
@maximocaceres4685 Linear algebra is a very tedious subject if done by hand. When using a computer it becomes much more manageable.
Ggs
Alright, let me know if you have a question about absolute value equations. Good luck with your studies :)
Thanks
@@mr_qualityy You are welcome! :)