Question

Can someone explain how to do this? |x - 6| > 2

Answer 1

i can

Answer 2

(I think you're supposed to do it now, jasonsmart ;))

Answer 3

Then please do explain.

Answer 4

oh I GOT THIS!!

Answer 5

|x| > a

means

x > a

or

x<-a

where a is nonnegative

Answer 6

so what you do .......YOU SON A MOTHER!!!!

Answer 7

I did not understand a thing....

Answer 8

yes my time to shine!

Answer 9

al right so what you do is add six to the other side of the greater sign and it will be X>2+6
so the answer would be X<8

Answer 10

4<x V x>8

Answer 11

....

Answer 12

jason I think you're wrong. If the solution is x<8 then, for example |6-6|>2 is not true. My solution is right!

Answer 13

whut....

Answer 14

Okay... Come on, I need to know how to solve these equations... So, I need a correct explanation.

Answer 15

.....you dare condescend me!.....

Answer 16

.... well you didnt give a very god question

Answer 17

|x-6|>2
You have to do x-6>2 and x-6<-2
First is x>8 and second is x<4

Right?

Answer 18

Achab is right.

Answer 19

whut....?

Answer 20

I don't understand his explanation....

Answer 21

I'll explain you now.

Answer 22

....

Answer 23

\[\Large x \in ( - \infty ,4) \cup (8,\infty )\]

Answer 24

Basically TheHero, the absolute value of a number is the positive form of that number. |-4| = 4, |-199123| = 199123, |12| = 12 and so on.

This question means that if you take 6 away from x, then the absolute result must be greater than two. If x is greater than 4 but less than 8 - say, 5 - then you'll end up with |5-6| = 1, which is not greater than 2, so it is invalid. But if you take a value less than 4 - say, 2 - then you end up with |2-6| = 4 which is greater than 2 so is valid. Similarly, over 8 - say, 15 - you get |15-6| = 9 which is greater than 2 and also valid.

Answer 25

we are solving for x?

Answer 26

I think i'm better off posting the questions.... Thanks anyways...

Answer 27

sorry i tried...

Answer 28

It's okay~ , you did good c:

Answer 29

yaey

Answer 30

\[|x-6|>2\] means that you have to search which x, subtracted from 6 and Absolutized, are higher than 2 (2,01; 2,00000001; 3;4, etc.).
You have to do 2 disequations.
The first one you have to do: x-6>2. Only remove the | | and solve it. It will be x>8, right?

Well, for the second you have to do: x-6<-2
You have to change the > to < and change the 2 to -2 (change the sign).
Now x<4 right?

Well, in the first you found x>8, in the second x<4, then, the xs are under 4 and higher than 8, right?
The solution is:
\[x<4 \] V (and) \[x>8\].
Prove yourself that if you take a x lower than 4 or higher than 8 it'll be correct.
Hope you've understand!