4(3y+1)=12(y-2)+27

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Solution for 4(3y+1)=12(y-2)+27 equation: 



4(3y+1)=12(y-2)+27

We move all terms to the left:

4(3y+1)-(12(y-2)+27)=0

We multiply parentheses

12y-(12(y-2)+27)+4=0



We calculate terms in parentheses: -(12(y-2)+27), so:

12(y-2)+27

We multiply parentheses

12y-24+27

We add all the numbers together, and all the variables

12y+3

Back to the equation:

-(12y+3)



We get rid of parentheses

12y-12y-3+4=0

We add all the numbers together, and all the variables

1!=0

There is no solution for this equation

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