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(1/3)(n+1)=(1/6)(3n-5)
We move all terms to the left:
(1/3)(n+1)-((1/6)(3n-5))=0
Domain of the equation: 3)(n+1)!=0
n∈R
Domain of the equation: 6)(3n-5))!=0We add all the numbers together, and all the variables
n∈R
(+1/3)(n+1)-((+1/6)(3n-5))=0
We multiply parentheses ..
(+n^2+1/3*1)-((+1/6)(3n-5))=0
We calculate fractions
((+n^2+1*6)(3n-5)))/18n^2+()/18n^2=0
We can not solve this equation
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