(4y+1)/3=(y+3)/2

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Solution for (4y+1)/3=(y+3)/2 equation: 



(4y+1)/3=(y+3)/2

We move all terms to the left:

(4y+1)/3-((y+3)/2)=0

We calculate fractions


4y/()+(-((y+3)*3)/()=0



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

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

We multiply all the terms by the denominator


-((y+3)*3)



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

(y+3)*3

We multiply parentheses

3y+9

Back to the equation:

-(3y+9)



We get rid of parentheses

-3y-9

Back to the equation:

+(-3y-9)



We get rid of parentheses

4y/()-3y-9=0

We multiply all the terms by the denominator


4y-3y*()-9*()=0

We add all the numbers together, and all the variables

4y-3y*()=0

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