anonymous
  • 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 :-)
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
ok, let's see
anonymous
  • anonymous
\[x-2[x-3(x+4)-5] = 3(2x-[x-8(x-4)])-2\] wow, lots of groups!
anonymous
  • anonymous
ok, so let's think about what the goal is, the goal is unite the groups on each side then solve for x

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anonymous
  • anonymous
yes
anonymous
  • anonymous
x=6
anonymous
  • anonymous
??? that fast?
anonymous
  • anonymous
adfriedman, nice but so fast!
anonymous
  • anonymous
it was just expanding brackets
anonymous
  • anonymous
right, ok fastff, a good approach could be to work from the inside out
anonymous
  • anonymous
focus on the left side: x−2[x−3(x+4)−5]
anonymous
  • anonymous
so let's start in the middle of this with 3(x+4)...can you expand this please?
anonymous
  • anonymous
to 3x+12
anonymous
  • anonymous
nice. so the equation is now x−2[x−(3x+12)−5] = x-2[x-3x-12-5]
anonymous
  • anonymous
the really quick approach is to note that only constants multiply the bracketed terms
anonymous
  • anonymous
ok, now simplify the expression with the [ ... ]
anonymous
  • anonymous
-3x+12x-5?
anonymous
  • anonymous
or -3x-12x-5?
anonymous
  • anonymous
-15x-5?
anonymous
  • anonymous
almost...forgot about the x
anonymous
  • anonymous
so you have x-3x-12-5....ok
anonymous
  • anonymous
simplify this: start with the x-terms
anonymous
  • anonymous
aren't we suppose to distribute?
anonymous
  • anonymous
distribute what?
anonymous
  • anonymous
so inside the [ ] you have x-3x-12-5 which becomes -2x-17, right?
anonymous
  • anonymous
fastff, how's it going?

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