How do you solve #18=3(3x-6)#?

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Answer 1 · 2016-07-31T14:39:38

#x = 4#

There are two methods:

Method 1. Divide both sides by 3.

#18/3 = (3(3x-6))/3 " "rArr cancel18^6/cancel3 = (cancel3(3x-6))/cancel3#

#6 = 3x-6#

#12 = 3x#

#x=4#

Method 2: Multiply 3 into the bracket:

#18 = 3(3x-6)#

#18 = 9x -18#

#36 = 9x#

#x = 4#

The solutions are of course the same, but one method involves slightly larger numbers.