1/6(2x+12)=1/3(3x-2)

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Solution for 1/6(2x+12)=1/3(3x-2) equation:



1/6(2x+12)=1/3(3x-2)
We move all terms to the left:
1/6(2x+12)-(1/3(3x-2))=0
Domain of the equation: 6(2x+12)!=0
x∈R
Domain of the equation: 3(3x-2))!=0
x∈R
We calculate fractions
(3x3/(6(2x+12)*3(3x-2)))+(-6x2/(6(2x+12)*3(3x-2)))=0
We calculate terms in parentheses: +(3x3/(6(2x+12)*3(3x-2))), so:
3x3/(6(2x+12)*3(3x-2))
We multiply all the terms by the denominator
3x3
We add all the numbers together, and all the variables
3x^3
We do not support expression: x^3

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