(n+5)/6=2+(2n+3)/3

Solution for (n+5)/6=2+(2n+3)/3 equation:

(n+5)/6=2+(2n+3)/3

We move all terms to the left:

(n+5)/6-(2+(2n+3)/3)=0

We calculate fractions


n/()+(-(2+(2n+3)*6)/()=0


We calculate terms in parentheses: +(-(2+(2n+3)*6)/(), so:

-(2+(2n+3)*6)/(

We multiply all the terms by the denominator


-(2+(2n+3)*6)


We calculate terms in parentheses: -(2+(2n+3)*6), so:

2+(2n+3)*6

determiningTheFunctionDomain
(2n+3)*6+2

We multiply parentheses

12n+18+2

We add all the numbers together, and all the variables

12n+20

Back to the equation:

-(12n+20)


We get rid of parentheses

-12n-20

Back to the equation:

+(-12n-20)


We get rid of parentheses

n/()-12n-20=0

We multiply all the terms by the denominator


n-12n*()-20*()=0

We add all the numbers together, and all the variables

n-12n*()=0