(3y-2)/3+(2y+3)/3=(y+7)/6

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Solution for (3y-2)/3+(2y+3)/3=(y+7)/6 equation: 



(3y-2)/3+(2y+3)/3=(y+7)/6

We move all terms to the left:

(3y-2)/3+(2y+3)/3-((y+7)/6)=0

We calculate fractions


5y/()+(-((y+7)*3)/()=0



We calculate terms in parentheses: +(-((y+7)*3)/(), so:

-((y+7)*3)/(

We multiply all the terms by the denominator


-((y+7)*3)



We calculate terms in parentheses: -((y+7)*3), so:

(y+7)*3

We multiply parentheses

3y+21

Back to the equation:

-(3y+21)



We get rid of parentheses

-3y-21

Back to the equation:

+(-3y-21)



We get rid of parentheses

5y/()-3y-21=0

We multiply all the terms by the denominator


5y-3y*()-21*()=0

We add all the numbers together, and all the variables

5y-3y*()=0

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