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Mathematics 16 Online
OpenStudy (anonymous):

What is the solution set for |x – 2|– 1 > 2? x > 5 or x < –1 x < 5 or x > –1 x > –5 or x < 1 x < 5 or x > –5

OpenStudy (anonymous):

first add 1 to both sides

OpenStudy (anonymous):

ok hold on

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

what did u get?

OpenStudy (anonymous):

um.........hold on

OpenStudy (anonymous):

|x – 2|– 1 > 2 +1 +1

OpenStudy (anonymous):

like this?

OpenStudy (jiteshmeghwal9):

|x-2| -(x-2)=-x+2x (x-2)=x-2

OpenStudy (anonymous):

|x – 2|– 1 > 2 Ix-2I-1+1>2+1 Ix-2I>3

OpenStudy (jiteshmeghwal9):

x+2*

OpenStudy (anonymous):

Did u understand there?

OpenStudy (anonymous):

yea

OpenStudy (anonymous):

sort of

OpenStudy (anonymous):

Now, HINT: IF IxI>a then x<-a and x>a

OpenStudy (anonymous):

Does that help?

OpenStudy (anonymous):

x > –5 or x < 1

OpenStudy (anonymous):

answer!^^^

OpenStudy (anonymous):

BOOYA

OpenStudy (anonymous):

Nope

OpenStudy (anonymous):

awwww :(

OpenStudy (jiteshmeghwal9):

so one time -x+2-1>2 & one time x-2-1>2 solve this & gt ur answer :)

OpenStudy (anonymous):

thanks

OpenStudy (anonymous):

What did u get? @evergirl

OpenStudy (anonymous):

x > 5 or x < –1

OpenStudy (anonymous):

Yep

OpenStudy (anonymous):

Did u understand how?

OpenStudy (anonymous):

a little

OpenStudy (anonymous):

IF IxI>a then x<-a and x>a

OpenStudy (anonymous):

ya

OpenStudy (anonymous):

Now, Ix-2I>3 x-2<-3 AND x-2>3

OpenStudy (anonymous):

Got it?

OpenStudy (anonymous):

yesh

OpenStudy (anonymous):

GREAT

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