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

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



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

We move all terms to the left:

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

We get rid of parentheses

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



We calculate terms in parentheses: -(2(y+1)+4y), so:

2(y+1)+4y

We add all the numbers together, and all the variables

4y+2(y+1)

We multiply parentheses

4y+2y+2

We add all the numbers together, and all the variables

6y+2

Back to the equation:

-(6y+2)



We add all the numbers together, and all the variables

-3y-(6y+2)-25=0

We get rid of parentheses

-3y-6y-2-25=0

We add all the numbers together, and all the variables

-9y-27=0

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

-9y=27

y=27/-9

y=-3

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