3(b-3)=2(b+2)+21

Solution for 3(b-3)=2(b+2)+21 equation:

3(b-3)=2(b+2)+21

We move all terms to the left:

3(b-3)-(2(b+2)+21)=0

We multiply parentheses

3b-(2(b+2)+21)-9=0


We calculate terms in parentheses: -(2(b+2)+21), so:

2(b+2)+21

We multiply parentheses

2b+4+21

We add all the numbers together, and all the variables

2b+25

Back to the equation:

-(2b+25)


We get rid of parentheses

3b-2b-25-9=0

We add all the numbers together, and all the variables

b-34=0

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

b=34