How do you solve #(3x-4) / 6 + 1/3 =( 3x) / 2#?

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Answer 1 · 2016-01-11T00:47:26

#x=-1/3#

#(3x-4)/6+1/3-(3x)/2=0#

#LCD =6# (Lowest Common Denominator)

#(1*(3x-4)+2*1-3*3x)/6=0#

#3x-4+2-9x=0#

#-6x-2=0#

#-6x=2#

#6x=-2#

#x=-cancel(2)^1/cancel(6)^3=-1/3#