3(b-1)-8(b+4)=1-3b+b-6

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Solution for 3(b-1)-8(b+4)=1-3b+b-6 equation: 



3(b-1)-8(b+4)=1-3b+b-6

We move all terms to the left:

3(b-1)-8(b+4)-(1-3b+b-6)=0

We add all the numbers together, and all the variables

3(b-1)-8(b+4)-(-2b-5)=0

We multiply parentheses

3b-8b-(-2b-5)-3-32=0

We get rid of parentheses

3b-8b+2b+5-3-32=0

We add all the numbers together, and all the variables

-3b-30=0

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

-3b=30

b=30/-3

b=-10

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