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2/9+1/2=2/3(n+3)
We move all terms to the left:
2/9+1/2-(2/3(n+3))=0
Domain of the equation: 3(n+3))!=0We calculate fractions
n∈R
()/(3(n+3))*9*2)+(12nn/(3(n+3))*9*2)+(27nn/(3(n+3))*9*2)=0
We calculate terms in parentheses: +()/(3(n+3))*9*2)+(12nn/(3(n+3))*9*2)+(27nn/(3(n+3))*9*2), so:
)/(3(n+3))*9*2)+(12nn/(3(n+3))*9*2)+(27nn/(3(n+3))*9*2
We can not solve this equation
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