-4(b+1)+2b=3(b-1)

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Solution for -4(b+1)+2b=3(b-1) equation: 



-4(b+1)+2b=3(b-1)

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

2b-4b-(3(b-1))-4=0



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

3(b-1)

We multiply parentheses

3b-3

Back to the equation:

-(3b-3)



We add all the numbers together, and all the variables

-2b-(3b-3)-4=0

We get rid of parentheses

-2b-3b+3-4=0

We add all the numbers together, and all the variables

-5b-1=0

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

-5b=1

b=1/-5

b=-1/5

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