Mathematics
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OpenStudy (anonymous):
I have a very hard question (for myself...)
x-2[x-3(x+4)-5]=3{2x-[x-8(x-4)]}-2
please help :-)

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OpenStudy (anonymous):
ok, let's see

OpenStudy (anonymous):
\[x-2[x-3(x+4)-5] = 3(2x-[x-8(x-4)])-2\]
wow, lots of groups!

OpenStudy (anonymous):
ok, so let's think about what the goal is, the goal is unite the groups on each side then solve for x

OpenStudy (anonymous):
yes

OpenStudy (anonymous):
x=6

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OpenStudy (anonymous):
??? that fast?

OpenStudy (anonymous):
adfriedman, nice but so fast!

OpenStudy (anonymous):
it was just expanding brackets

OpenStudy (anonymous):
right, ok fastff, a good approach could be to work from the inside out

OpenStudy (anonymous):
focus on the left side:
x−2[x−3(x+4)−5]

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OpenStudy (anonymous):
so let's start in the middle of this with 3(x+4)...can you expand this please?

OpenStudy (anonymous):
to 3x+12

OpenStudy (anonymous):
nice. so the equation is now x−2[x−(3x+12)−5] = x-2[x-3x-12-5]

OpenStudy (anonymous):
the really quick approach is to note that only constants multiply the bracketed terms

OpenStudy (anonymous):
ok, now simplify the expression with the [ ... ]

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OpenStudy (anonymous):
-3x+12x-5?

OpenStudy (anonymous):
or -3x-12x-5?

OpenStudy (anonymous):
-15x-5?

OpenStudy (anonymous):
almost...forgot about the x

OpenStudy (anonymous):
so you have x-3x-12-5....ok

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OpenStudy (anonymous):
simplify this: start with the x-terms

OpenStudy (anonymous):
aren't we suppose to distribute?

OpenStudy (anonymous):
distribute what?

OpenStudy (anonymous):
so inside the [ ] you have x-3x-12-5 which becomes -2x-17, right?

OpenStudy (anonymous):
fastff, how's it going?