(n-1)/3=(2n+1)/9

Solution for (n-1)/3=(2n+1)/9 equation:

(n-1)/3=(2n+1)/9

We move all terms to the left:

(n-1)/3-((2n+1)/9)=0

We calculate fractions


n/()+(-((2n+1)*3)/()=0


We calculate terms in parentheses: +(-((2n+1)*3)/(), so:

-((2n+1)*3)/(

We multiply all the terms by the denominator


-((2n+1)*3)


We calculate terms in parentheses: -((2n+1)*3), so:

(2n+1)*3

We multiply parentheses

6n+3

Back to the equation:

-(6n+3)


We get rid of parentheses

-6n-3

Back to the equation:

+(-6n-3)


We get rid of parentheses

n/()-6n-3=0

We multiply all the terms by the denominator


n-6n*()-3*()=0

We add all the numbers together, and all the variables

n-6n*()=0