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

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



6(b+3)-3(b+2)=-3(2b+1)-3
We move all terms to the left:
6(b+3)-3(b+2)-(-3(2b+1)-3)=0
We multiply parentheses
6b-3b-(-3(2b+1)-3)+18-6=0
We calculate terms in parentheses: -(-3(2b+1)-3), so:
-3(2b+1)-3
We multiply parentheses
-6b-3-3
We add all the numbers together, and all the variables
-6b-6
Back to the equation:
-(-6b-6)
We add all the numbers together, and all the variables
3b-(-6b-6)+12=0
We get rid of parentheses
3b+6b+6+12=0
We add all the numbers together, and all the variables
9b+18=0
We move all terms containing b to the left, all other terms to the right
9b=-18
b=-18/9
b=-2

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