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

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



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

We move all terms to the left:

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



Domain of the equation: 3n!=0

n!=0/3

n!=0

n∈R





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

n∈R



We calculate fractions


((n-6)*(3n+1)))/12n^2+(-((n-5)*3n)/12n^2=0

We calculate fractions


(((n-6)*(3n+1)))*12n^2)/(12n^2+(*12n^2)+(-((n-5)*3n)*12n^2)/(12n^2+(*12n^2)=0



We calculate terms in parentheses: +(-((n-5)*3n)*12n^2)/(12n^2+(*12n^2), so:

-((n-5)*3n)*12n^2)/(12n^2+(*12n^2

We multiply all the terms by the denominator


-((n-5)*3n)*12n^2)+((*12n^2)*(12n^2

Back to the equation:

+(-((n-5)*3n)*12n^2)+((*12n^2)*(12n^2)



We get rid of parentheses

(((n-6)*(3n+1)))*12n^2)/(12n^2+*12n^2+(-((n-5)*3n)*12n^2)+((*12n^2)*12n^2=0

We multiply all the terms by the denominator


(((n-6)*(3n+1)))*12n^2)+(*12n^2)*(12n^2+((-((n-5)*3n)*12n^2))*(12n^2+(((*12n^2)*12n^2)*(12n^2=0

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