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1/3(x+1)+2=1/6(3x-5)
We move all terms to the left:
1/3(x+1)+2-(1/6(3x-5))=0
Domain of the equation: 3(x+1)!=0
x∈R
Domain of the equation: 6(3x-5))!=0We calculate fractions
x∈R
(6x3/(3(x+1)*6(3x-5)))+(-3xx/(3(x+1)*6(3x-5)))+2=0
We calculate terms in parentheses: +(6x3/(3(x+1)*6(3x-5))), so:
6x3/(3(x+1)*6(3x-5))
We multiply all the terms by the denominator
6x3
We add all the numbers together, and all the variables
6x^3
We do not support expression: x^3
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