(3y-5)/6=(2y-1)/3+2

Solution for (3y-5)/6=(2y-1)/3+2 equation:

(3y-5)/6=(2y-1)/3+2

We move all terms to the left:

(3y-5)/6-((2y-1)/3+2)=0

We calculate fractions


(9y-15)/()+(-((2y-1)*6)/()=0


We calculate terms in parentheses: +(-((2y-1)*6)/(), so:

-((2y-1)*6)/(

We multiply all the terms by the denominator


-((2y-1)*6)


We calculate terms in parentheses: -((2y-1)*6), so:

(2y-1)*6

We multiply parentheses

12y-6

Back to the equation:

-(12y-6)


We get rid of parentheses

-12y+6

Back to the equation:

+(-12y+6)


We get rid of parentheses

(9y-15)/()-12y+6=0

We multiply all the terms by the denominator


(9y-15)-12y*()+6*()=0

We add all the numbers together, and all the variables

(9y-15)-12y*()=0

We get rid of parentheses

9y-12y*()-15=0

We move all terms containing y to the left, all other terms to the right

9y-12y*()=15