Solving Advanced Absolute Value Inequalities

[@Greenemath]

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

[@hqs9585]

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]

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]

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

[@davegoodo3603]

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]

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

[@davegoodo3603]

14 mins and 20 seconds.

[@Greenemath]

@@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! 😎

[@videodetermined8485]

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

[@Greenemath]

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.

[@choongchoongkok537]

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]

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.

[@ckdjcj3091]

This method does not work for all of them

[@Greenemath]

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