8(b+2)-(3b-(b-4))=28

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Solution for 8(b+2)-(3b-(b-4))=28 equation: 



8(b+2)-3b-(b-4)=28

We move all terms to the left:

8(b+2)-3b-(b-4)-(28)=0

We add all the numbers together, and all the variables

-3b+8(b+2)-(b-4)-28=0

We multiply parentheses

-3b+8b-(b-4)+16-28=0

We get rid of parentheses

-3b+8b-b+4+16-28=0

We add all the numbers together, and all the variables

4b-8=0

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

4b=8

b=8/4

b=2

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