3(y+3)=13y+3913(y+3)=13y+39

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Solution for 3(y+3)=13y+3913(y+3)=13y+39 equation: 



3(y+3)=13y+3913(y+3)=13y+39

We move all terms to the left:

3(y+3)-(13y+3913(y+3))=0

We multiply parentheses

3y-(13y+3913(y+3))+9=0



We calculate terms in parentheses: -(13y+3913(y+3)), so:

13y+3913(y+3)

We multiply parentheses

13y+3913y+11739

We add all the numbers together, and all the variables

3926y+11739

Back to the equation:

-(3926y+11739)



We get rid of parentheses

3y-3926y-11739+9=0

We add all the numbers together, and all the variables

-3923y-11730=0

We move all terms containing y to the left, all other terms to the right

-3923y=11730

y=11730/-3923

y=-2+3884/3923

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