Mathematics 29 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

??? that fast?

OpenStudy (anonymous):

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]

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 [ ... ]

OpenStudy (anonymous):

-3x+12x-5?

OpenStudy (anonymous):

or -3x-12x-5?

OpenStudy (anonymous):

-15x-5?

OpenStudy (anonymous):

OpenStudy (anonymous):

so you have x-3x-12-5....ok

OpenStudy (anonymous):

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?

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