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

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



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

We move all terms to the left:

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



Domain of the equation: 2(y+2)!=0

y∈R





Domain of the equation: 3(2y+6))!=0

y∈R



We calculate fractions


(3y2/(2(y+2)*3(2y+6)))+(-2yy/(2(y+2)*3(2y+6)))=0



We calculate terms in parentheses: +(3y2/(2(y+2)*3(2y+6))), so:

3y2/(2(y+2)*3(2y+6))

We multiply all the terms by the denominator


3y2

We add all the numbers together, and all the variables

3y^2

Back to the equation:

+(3y^2)





We calculate terms in parentheses: +(-2yy/(2(y+2)*3(2y+6))), so:

-2yy/(2(y+2)*3(2y+6))

We multiply all the terms by the denominator


-2yy

Back to the equation:

+(-2yy)



We get rid of parentheses

3y^2-2yy=0

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