4(6y+3)+5(3y+1)=27

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Solution for 4(6y+3)+5(3y+1)=27 equation: 



4(6y+3)+5(3y+1)=27

We move all terms to the left:

4(6y+3)+5(3y+1)-(27)=0

We multiply parentheses

24y+15y+12+5-27=0

We add all the numbers together, and all the variables

39y-10=0

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

39y=10

y=10/39

y=10/39

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