2y-3+3(3y+2)=-7(y+1)

Solution for 2y-3+3(3y+2)=-7(y+1) equation:

2y-3+3(3y+2)=-7(y+1)

We move all terms to the left:

2y-3+3(3y+2)-(-7(y+1))=0

We multiply parentheses

2y+9y-(-7(y+1))+6-3=0


We calculate terms in parentheses: -(-7(y+1)), so:

-7(y+1)

We multiply parentheses

-7y-7

Back to the equation:

-(-7y-7)


We add all the numbers together, and all the variables

11y-(-7y-7)+3=0

We get rid of parentheses

11y+7y+7+3=0

We add all the numbers together, and all the variables

18y+10=0

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

18y=-10

y=-10/18

y=-5/9