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

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Solution for 7(b+2)+3(b-4)=4(b-1) equation: 



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

We move all terms to the left:

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

We multiply parentheses

7b+3b-(4(b-1))+14-12=0



We calculate terms in parentheses: -(4(b-1)), so:

4(b-1)

We multiply parentheses

4b-4

Back to the equation:

-(4b-4)



We add all the numbers together, and all the variables

10b-(4b-4)+2=0

We get rid of parentheses

10b-4b+4+2=0

We add all the numbers together, and all the variables

6b+6=0

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

6b=-6

b=-6/6

b=-1

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