Inequalites involving absoulte values [resolved]

[Dillinger4]

Help~* I happened to see this on a test i have to take next week so i thought that i would try and learn how to do these but i am having trouble.

For a postive real number a |x| < a breaks into -a < x and x < a which can be treated as a single compound inequality -a<x<a

|x| >a breaks into x < -a or x > a.

Now i just saw two problems which i donot understand.
|4x+2| < 6 // are they saying |x| < a ?
|2x - 1| > 3 // are they saying that |x| > a?

I think that the defination of an absoulte value that i had i my mind is wrong. Isn't the absoulte value of a postive number itself and the absoulte value of a negative number it's positive complement? So in this case how can |4x+2| < 6 be correct?

Thanks for any help guys.

[HarryW]

You have this inequality then, and you have to find the domain of x, is that right?

|4x+2| < 6

I forgot the terminology a bit, domain may be the wrong word.

Okay, well that means:

-6 < 4x +2

and

4x +2 < 6

so :

-6 - 2 < 4x

-8 < 4x

-2 < x <---- one end of the range

and:

4x +2 < 6

4x < 6 - 2

4x < 4

x < 1 <---- other end of the range


so the domain (if that's the right word) is :

-2 < x < 1


[edited for a stupid mistake]

[NotLKH]

Originally posted by Dilenger4
[BNow i just saw two problems which i donot understand.
|4x+2| < 6 // are they saying |x| < a ?
|2x - 1| > 3 // are they saying that |x| > a?
[/B]

Basically, if the Q is to plot all X that satisfy the eq: |4x + 2| < 6,
then it works out like this.

First of all, understand that |+-6| = 6.

So, for |4x + 2| = 6 then 4x + 2 = 6 or 4x + 2 = -6.
This is important, because we need to find the hotspots of x.

So, to start with, find the value for x when 4x + 2 = 6.
this would be x = 1.

There is No Other x which makes 4x + 2 = 6.
But since we're dealing with absolutes, there is the other eq.

Find the value of x such that 4x + 2 = -6.
This would be x = -2.

Alrighty, we now know the hot spots for x, 1 and -2.

But, since the eq IS NOT |4x + 2| = 6, but is really |4x + 2| < 6,
we have to figure out if we are going to shade all the values of x < 1 and x > -2,
or are we going to shade the values of x > 1 and x < -2.

Simply plug in a test value of x = 0 {since 0 is between 1 and -2},
and evaluate the validity of the eq |4x + 2| < 6.

So, is |4*0 + 2| < 6? Why, yes it is. So, the solution is all the values of x, such that -2 < x < 1.

Try working out |2x - 1| > 3 this way.

-Lou

[Dillinger4]

You have this inequality then, and you have to find the domain of x, is that right?

I forgot the terminology a bit, domain may be the wrong word.

Yes i think offhand that it's called an interval.

Hey thanks for replying guys. I worked out the problem Lou
and im not sure if im on the right track.

|2x - 1| > 3

Now we want to figure out:

2x - 1 = 3

Ok then x should be equal to 2.

2x - 1 = -3

Ok then x should be equal to -1

Alrighty, we now know the hot spots for x, 2 and -1.

I would have called it the G-Spot myself.

Simply plug in a test value of x = 1 {since 1 is between 2 and -1},
and evaluate the validity of the eq |2x- 1| > 3.

I don't think that this is it because |2.1- 1| > 3 is not greater than 3. And the only other number i can test it with is 2 which would not work either.

[NotLKH]

Originally posted by Dilenger4

Simply plug in a test value of x = 1 {since 1 is between 2 and -1},
and evaluate the validity of the eq |2x- 1| > 3.

The "since" part isn't really needed. All you want to do is test a value.

I would have chosen 0.

Right now, we know that the hotspots are 2 and -1. We want to know the reqion where |2*x - 1| > 3.
So, we know either -1 < x < 2, or the other; x > 2, x < -1.

So, lets test 0. if 0 when applied to the eq produces a value > 3, then we know -1 < x < 2, since 0 is within that range. Otherwise, if, when useing 0 in the eq, it produces a value < 3, then we know it MUST be the other range of X that is valid, ie... x > 2 & x < -1.

So, what do we get with 0?

|2 * 0 - 1| => |0 - 1| => |-1| => 1. And 1 is NOT > 3. so, the answer for x MUST be x > 2 and x < -1.

-Lou

[Dillinger4]

I took this |x - 6 | > 10 and figured it out based on x<-a or x > a
x-6 + 6 > 10 + - 6

So x > 16 or x < -4

But i honestly don't see how im going to remember the formulas
when i see the problem. I mean if i see |x - 6 | > 10 i have to remember x<-a or x > a and if i see |x - 6 | < 10 i have to remember -a<x<a.

[Dillinger4]

I tried another one i just made up. |x+9|> 26
x+9 + (-9) > 26 + 9 or (-9)

So x > 35 or x < 17

Dose this look right? It seems easy to do it this way as long as that answers im getting are in fact correct.

[NotLKH]

I see. You're trying to "Balance" the equation to a solution.
Heh, I had a hard time Bending my mind around "Ok, If I subtract a Number on One side, If got to both Add and Subtract it on the other... Or something like that! ".

Are you EXPECTED to do it this way? or can you just give the solution from any method you like.

BTW, How do you "Balance" the following and get the ranges of x?

|x^2 - 30*x + 9| < 7

I think finding the 2 sets of Hot Spots for x, or the 2 sets of roots for the eq's x^2 - 30*x + 2 = 0, and x^2 - 30*x + 16 = 0
is easier.

-Lou

[NotLKH]

Originally posted by Dilenger4
I tried another one i just made up. |x+9|> 26
x+9 + (-9) > 26 + 9 or (-9)

So x > 35 or x < 17

Dose this look right? It seems easy to do it this way as long as that answers im getting are in fact correct.

Test it.

x > 35? lets try 36.
|36 + 9| = |45| > 26. Yep, it is.

However, so is 34, 33, ...
and 34, 33, is < 35.
So,,, Your solution isn't quite right.

in Actuality x > 17, and x < -35.

My Way:

|x + 9| > 26
So, Find the "HotSpots", limits, roots, whatever you want to call it, for when
|x + 9| = 26.

Extract #1) Where x + 9 = 26.
::::x = 26 - 9, or x = 17.

Extract #2) Where x + 9 = -26.
::::x = -26 - 9, or x = -35.

So, the HotSpots are -35 and 17.
We know x can be in either of 2 ranges, for a linear equation.
either (x < -35 and x > 17) or (x > -35 and x < 17).

Lets test 0.
First of all, 0 > -35, and also 0 < 17. so if 0 solves the eq., then we know x > -35 and x < 17.

So, is |0 + 9| > 26?

Nope. Its Less than 26. so x Cannot be between -35 and 17.
So x must be OUTSIDE the range of -35 and 17.

So, X < -35, and X > 17.

heh, this is fun!!!

[Dillinger4]

For this problem |x-6| > 10
x-6 + 6 > 10 + -6
x > 16 or x < -4

So i fugured i could do the same with |x + 9| > 26
x 9 + (-9) > 26 + -9 0r 9

x > 35 or x < 17

But i guess it's not possible.

[NotLKH]

Sorry, I was in Edit during your last post. See what I've added.

-Lou

[Dillinger4]

Ahhhh ok. So for |x+13| > 28

? + 13 = 28
15 + 13 = 28
so x= 15

? + 13 = -28
-41 + 13 = 28
so x = -41

(x < 28 and x > -41) or (x > 28 and x < -41)

[Dillinger4]

Now that i have (x < 28 and x > -41) or (x > 28 and x < -41)
Let's test with 0 like you did. |0 + 13| > 28 Nope, so it's outside range.

So the answer must be x < 28 and x > -41

[NotLKH]

No.
You were almost right.

Let me correct what you posted:
Ahhhh ok. So for |x+13| > 28

? + 13 = 28
15 + 13 = 28
so x= 15 <---Good!

? + 13 = -28
-41 + 13 = -28 <---Lost your neg 28
so x = -41 <---------Good

(x < 15 and x > -41) or (x > 15 and x < -41) <----You had 28 instead of 15!!!! Watch Out!!! Go thru your calc's at least 2 times, once on paper, then Line By Line, Mentally think what your next line is supposed to be, before you look at the next line.

__________________________
Ok, Now that we have the correct ranges,



Test it. You know that x could be 1 of those 2 ranges. You just don't know which range yet, for certainty.

Try 0. First of all, which range is 0 included in.

Then, Plug it into the eq. If the eq turns out valid, then that range is valid. If it turns out the eq, after plugging 0 in, is Invalid, then the range which 0 is part of is Not the solution. So, it must be the other range.

[Dillinger4]

I think i have it now. |x + 43| > 96

x+ 43 = 96
x = 53

x+ 43 = -96
x = -139

So (x < 53 and x > -139) or (x > 53 and x < -139)

So the answer is (x > 53 and x < -139)

This seems to apply when using "|x| >a breaks into x < -a or x > a." Now i want to see if it applies the other way.

[NotLKH]

-Lou

[Dillinger4]

Hey thanks Lou and Harry. Can i buy you guys somthing for Christmas!? Maybe a nice fruit basket?

[Dillinger4]

A beer at the Winfield, in Rochester NY wouldn't be unappreciated!

That's pretty cool you live in Rochester NY. I live in NJ right outside of the GWB. I uesd to live in Hartsadale NY right outside of WhitePlanes. But anyway....... thanks again. Oh and ill have too keep the points you brought up about using 0 or a hot spot for the testing process. Next on my list is quadratic equations.