You flip the direction of the inequality symbol when you multiply or divide both sides by a negative. Distributing a negative does not change the direction of the inequality symbol.
@Jovan Rio Tjandra that goes back to the definition of absolute value. It's actually explained at the beginning of each problem. Where |u| > k then u < -k or u > k and then for the next one we have |u| < k, which is -k < u < k. So one is a greater than while the other is a less than. You just need to think about the concept of absolute value. You may want to start with the more basic absolute value inequalities. Which was back in Lesson 44. greenemath.com/College_Algebra/44/Solving-Absolute-Value-Inequalities.html I have a video coming out soon (Saturday) where all the absolute value problems will be put together as one. You may want to watch that as you may fill in some gaps.
Thanks for the great video! I'd appreciate if you could answer a question: Why doesn't it work to use the interval method (as seen on problem 5) to solve |3x-2| - |-(x/3) +1| =0 ? I did it and the results would seem ok but none of them belongs to their respective intervals. Am I doing something wrong? I got 9/10 for the interval x
This is a solution video for a practice test. Watch the video lesson if you want a full explanation of why we are setting up intervals on the number line.
I have no idea what you are trying to do. There isn't an equality symbol or inequality symbol. Also, you have two left absolute value bars and only one on the right.
Practice Test:
greenemath.com/College_Algebra/72/Absolute-Value-Inequalities-2PracticeTest.html
These questions are all I need. Thanks a lot and I have a suggestion. It will be better if you write 2's better.
My handwriting has always been hot garbage.
Many thanks this is just what I needed.
You are welcome.
Why do you not flip the inequality when you are distributing negatives to absolute values?
You flip the direction of the inequality symbol when you multiply or divide both sides by a negative. Distributing a negative does not change the direction of the inequality symbol.
Woooh Thanks alots Sir for your splendid presentation,,,,it is only full of vital concepts
You are most welcome
When do we have to think of the overlap and when can we just take all the regions that are valid?
Is that a particular problem that you are lost on? There are a few different set ups in this video.
@@Greenemath For example in 9 and 10, both questions are similar in style. In 9 you take everything, but for 10 you take the overlap.
@Jovan Rio Tjandra that goes back to the definition of absolute value. It's actually explained at the beginning of each problem. Where |u| > k then u < -k or u > k and then for the next one we have |u| < k, which is -k < u < k. So one is a greater than while the other is a less than. You just need to think about the concept of absolute value. You may want to start with the more basic absolute value inequalities. Which was back in Lesson 44.
greenemath.com/College_Algebra/44/Solving-Absolute-Value-Inequalities.html
I have a video coming out soon (Saturday) where all the absolute value problems will be put together as one. You may want to watch that as you may fill in some gaps.
Everything i was looking for :)
Great, glad to hear that!
Thanks for the great video! I'd appreciate if you could answer a question: Why doesn't it work to use the interval method (as seen on problem 5) to solve |3x-2| - |-(x/3) +1| =0 ? I did it and the results would seem ok but none of them belongs to their respective intervals. Am I doing something wrong? I got 9/10 for the interval x
This is a solution video for a practice test. Watch the video lesson if you want a full explanation of why we are setting up intervals on the number line.
@@Greenemath Thanks a lot!!
Thank you
You're welcome
i'm so grateful
Glad to hear that!
helped a lot thankyou!!!!
You are very welcome!
thank you so much!!!!
You're welcome!
Danke sehr🙏 thank you, sir
You are very welcome.
Great video , thank you
Glad you enjoyed it
Graphing, if possible, takes away much of the mystery
It definitely helps on difficult problems.
How about arcsin(|x-1|-2x+3|-3)
I have no idea what you are trying to do. There isn't an equality symbol or inequality symbol. Also, you have two left absolute value bars and only one on the right.