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

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



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

We move all terms to the left:

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

We calculate fractions


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



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

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

We multiply all the terms by the denominator


-((3y-7)*3)



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

(3y-7)*3

We multiply parentheses

9y-21

Back to the equation:

-(9y-21)



We get rid of parentheses

-9y+21

Back to the equation:

+(-9y+21)



We get rid of parentheses

2y/()-9y+21=0

We multiply all the terms by the denominator


2y-9y*()+21*()=0

We add all the numbers together, and all the variables

2y-9y*()=0

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