3(1-b)-2(b+1)=9(b-1)+6

Solution for 3(1-b)-2(b+1)=9(b-1)+6 equation:

3(1-b)-2(b+1)=9(b-1)+6

We move all terms to the left:

3(1-b)-2(b+1)-(9(b-1)+6)=0

We add all the numbers together, and all the variables

3(-1b+1)-2(b+1)-(9(b-1)+6)=0

We multiply parentheses

-3b-2b-(9(b-1)+6)+3-2=0


We calculate terms in parentheses: -(9(b-1)+6), so:

9(b-1)+6

We multiply parentheses

9b-9+6

We add all the numbers together, and all the variables

9b-3

Back to the equation:

-(9b-3)


We add all the numbers together, and all the variables

-5b-(9b-3)+1=0

We get rid of parentheses

-5b-9b+3+1=0

We add all the numbers together, and all the variables

-14b+4=0

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

-14b=-4

b=-4/-14

b=2/7