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\[-3|x-2|=2\]
now we square this..
the only way to get rid of the absoute sign is to square it

or make two equations out of it.

ok wait spuare the hole thing

yes see the above simplified orm

\[\left[-3|x-2|\right]^2=2^2\]

ok wat about the + 10

I subtracted it from the two sides

notice that the right side became 12-10=2

where did u get the 12 frm

where did u get the 12 from

\[-3|x-2|+10=12\qquad\text{....given}\\
\qquad\qquad-10\quad-10
\]

ok srry im just learning this ok and then gives u this
[−3|x−2|]^2=2^2

that is ok.
well, this first gives
\[-3|x-2|=2\]
so, we the square it like you just wrote above

yes

\[(-3)^2(x-2)^2=(2^2)\]
notice that the absolute sign has disappeared

so 9*2x-4=4

how did u get 3/2

my bad... \[\Large{|x-2|=\color{red}{-}{2\over3}}\]

ooh its ok but how did u get that

divide the two sides by "3"

ok

\[{-3|x-2|\over-3}={2\over-3}\]

this gives the above form..
now, by definition of ABSOLUTE value, can it ever become negative?

HEY STOP STOP STOP I WROTE THE PROBLEM DWN WRONG
SRRY
THIS IS THE PROBLEM −3|2x + 6| = −12

ok...
you mean no "10" in there?

yea

this is so much different than what you first asked for

ok.. proceeding similarly, get the absolute quantity on its own...

ik b/c its 2 differnt problems in one

thanks

did you understand?

yes