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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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


We calculate terms in parentheses: -(3(-3b+1)+6(b+1)), so:

3(-3b+1)+6(b+1)

We multiply parentheses

-9b+6b+3+6

We add all the numbers together, and all the variables

-3b+9

Back to the equation:

-(-3b+9)


We get rid of parentheses

3b-9+3=0

We add all the numbers together, and all the variables

3b-6=0

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

3b=6

b=6/3

b=2