-28-(2y+2)=3(y+2)+4y

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Solution for -28-(2y+2)=3(y+2)+4y equation: 



-28-(2y+2)=3(y+2)+4y

We move all terms to the left:

-28-(2y+2)-(3(y+2)+4y)=0

We get rid of parentheses

-2y-(3(y+2)+4y)-2-28=0



We calculate terms in parentheses: -(3(y+2)+4y), so:

3(y+2)+4y

We add all the numbers together, and all the variables

4y+3(y+2)

We multiply parentheses

4y+3y+6

We add all the numbers together, and all the variables

7y+6

Back to the equation:

-(7y+6)



We add all the numbers together, and all the variables

-2y-(7y+6)-30=0

We get rid of parentheses

-2y-7y-6-30=0

We add all the numbers together, and all the variables

-9y-36=0

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

-9y=36

y=36/-9

y=-4

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