3(3b-3)-18=2(b+9)+11

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Solution for 3(3b-3)-18=2(b+9)+11 equation: 



3(3b-3)-18=2(b+9)+11

We move all terms to the left:

3(3b-3)-18-(2(b+9)+11)=0

We multiply parentheses

9b-(2(b+9)+11)-9-18=0



We calculate terms in parentheses: -(2(b+9)+11), so:

2(b+9)+11

We multiply parentheses

2b+18+11

We add all the numbers together, and all the variables

2b+29

Back to the equation:

-(2b+29)



We add all the numbers together, and all the variables

9b-(2b+29)-27=0

We get rid of parentheses

9b-2b-29-27=0

We add all the numbers together, and all the variables

7b-56=0

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

7b=56

b=56/7

b=8

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