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- anonymous

lxl-2=3

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Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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- anonymous

lxl-2=3

- katieb

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- anonymous

x = +- 5

- anonymous

lol lx+12l =8

- anonymous

I dont know how to do the absolute value symbols, so bear with me

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- anonymous

|x+12| = 8 is the same as | x -(-12)| = 8, therefore x is either -20 or -4.

- anonymous

simply, |a - b| is the distance between a and b on the real axis.

- anonymous

|6-x|=10
I know one of them is 4 but whats the other T_T

- anonymous

\[|x+12|=8\] solve two equations
\[x+12=8\] or
\[x+12=-8\] first one gives
\[x=-4\]
second one gives
\[x=-20\]

- anonymous

lol fail. I forgot that was a negative

- anonymous

lxl-3=6

- anonymous

\[|6-x|=10\] same as
\[|x-6|=10\]
solve
\[x-6=10\] or
\[x-6=-10\]

- anonymous

convince yourself that
\[|a-b|=|b-a|\] so you do not have to deal with the annoying
\[|6-x|\] just change to
\[|x-6|\]

- anonymous

whered you get the 10?

- anonymous

you wrote
\[|6-x|=10\] i just changed it to
\[|x-6|=10\]

- anonymous

lol.
Thought we were working on this one D:
lxl-3=6

- anonymous

this one is easier. this one is
\[|x|=9\]
making
\[x=\pm 9\]

- anonymous

okay for lx+4l=3 you would make it 4+x=-3 and 4=x=3, correct?

- anonymous

that last part was supposed to say 4+x=3

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