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There we are. We MUST understand this principle. |x| = x if x >= 0 Example |3| = 3 Absolute value does nothing to positive values |x| = -x if x < 0 Example |-3| = -(-3) = 3 Absolute value changes the sign of positive values
So, 2|x -5| < 32 Divide by 2, just to simplify. |x-5| < 16 Where is x-5 > 0.
don't we add 5 and x=5 or no
x-5 > 0 everywhere x > 5. Here's the really big step... If we PROMISE only to think about x > 5, our problem becomes: (x-5) < 16 because absolute values don't do anything to positive values.
that answer isn't on my paper they have:-6
2, and x<-11 or x>21
Why are we looking at answers? We're trying to understand the thinking. We don't yet have an answer. Did it makes sense so far? If we PROMISE only to think about x > 5, our problem becomes: (x-5) < 16 because absolute values don't do anything to positive values.
oh ok and yes I get it.
And the next piece, If we PROMISE only to think about x < 5, our problem becomes: -(x-5) < 16 because absolute values just change the sign of negative values. Still making sense?
I'm following you
Here we are: If x >= 5, solve x - 5 < 16 (Sorry, I forgot to include "=" up above.) If x < 5, solve -(x - 5) < 16 Can you solve those?
I think x<=21
Keep the IF statements. If x >= 5, then x < 21 If x < 5, then -x+5 < 16 or -x < 11 or x > -11 This is another important decision point. We don't quite have an answer.
oh okay still following
Almost done. If \(x \ge 5\) and \(x < 21\), we have \(5 \le x < 21\) as part of the solution. If you see why this is so, do the other piece just the same way.
what does ge and le stand for so I can do the problem I understand everything else but that
Oh, maybe your math isn't working. "ge" is greater than or equal to and "le" is less than or equal to.
okay I lost myself in my own work so far I have If \(x \le 5)and\ (-x \le 11) so (5\ge x\< -11) I think
There should be no 'e' in this set. If x < 5 and x > -11, we have -11 < x < 5 as a piece of the solution.
oh ok I think i got my answer x<-11 or x>21
Careful. We have two solution pieces. We must join them properly: 5≤x<21 or -11
oh I was looking at the wrong chose thanks you've helped me a lot with Algebra II
To be honest, what we just dragged through is pretty tedious. There are faster ways to solve some problems. What I hoped to do here was introduce you to the full fundamental basis - a method that will ALWAYS work no matter what else happens. Find faster ways, but don't forget the fundamentals.
3 | x + 5 | < 21
Fast way on this one: 3|x+5|<21 |x+5| < 7 This is kind of magic -7 < (x+5) < 7 Subtract 5 from all three pieces -12 < x < 2 Done.
you taught me something I didn't know at all and it seems easy once you got the hang of it thanks a lot
oh that's easy and quick 1 question were did you get the 1st < sign from
These two were very similar because the two pieces came back together in the end. It resulted in a contiguous solution. Don't try that with other problems. The fundamental way will always work, but the super quick and fancy way sometimes struggles.
In the first problem, we split it as x = 5 and it eventually came back together. Had we split up the second one, we would have split it at x = -5 and then it would have come back together in the end.
oh now I get it a little well I have another one that's tripping me up | x + 1| < 5
Same as the last one. |x+1| < 5 Is the same as -5 < (x+1) < 5 It's kind of a transformation you should memorize.
thanks @tkhunny I got a 100