2y/7=(y+3)/4

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Solution for 2y/7=(y+3)/4 equation: 



2y/7=(y+3)/4

We move all terms to the left:

2y/7-((y+3)/4)=0

We calculate fractions


8y^2/()+(-((y+3)*7)/()=0



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

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

We multiply all the terms by the denominator


-((y+3)*7)



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

(y+3)*7

We multiply parentheses

7y+21

Back to the equation:

-(7y+21)



We get rid of parentheses

-7y-21

Back to the equation:

+(-7y-21)



We get rid of parentheses

8y^2/()-7y-21=0

We multiply all the terms by the denominator


8y^2-7y*()-21*()=0

We add all the numbers together, and all the variables

8y^2-7y*()=0

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