1/6(3n-5)=1/3(n+1)

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Solution for 1/6(3n-5)=1/3(n+1) equation: 



1/6(3n-5)=1/3(n+1)

We move all terms to the left:

1/6(3n-5)-(1/3(n+1))=0



Domain of the equation: 6(3n-5)!=0

n∈R





Domain of the equation: 3(n+1))!=0

n∈R



We calculate fractions


(3nn/(6(3n-5)*3(n+1)))+(-6n3/(6(3n-5)*3(n+1)))=0



We calculate terms in parentheses: +(3nn/(6(3n-5)*3(n+1))), so:

3nn/(6(3n-5)*3(n+1))

We multiply all the terms by the denominator


3nn

Back to the equation:

+(3nn)





We calculate terms in parentheses: +(-6n3/(6(3n-5)*3(n+1))), so:

-6n3/(6(3n-5)*3(n+1))

We multiply all the terms by the denominator


-6n3

We add all the numbers together, and all the variables

-6n^3

We do not support enpression: n^3

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