b-(3-(b-(2-b)+4))=2b+6

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Solution for b-(3-(b-(2-b)+4))=2b+6 equation: 



b-(3-(b-(2-b)+4))=2b+6

We move all terms to the left:

b-(3-(b-(2-b)+4))-(2b+6)=0

We add all the numbers together, and all the variables

b-(3-(b-(-1b+2)+4))-(2b+6)=0

We get rid of parentheses

b-(3-(b-(-1b+2)+4))-2b-6=0



We calculate terms in parentheses: -(3-(b-(-1b+2)+4)), so:

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

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



We calculate terms in parentheses: -(b-(-1b+2)+4), so:

b-(-1b+2)+4

We get rid of parentheses

b+1b-2+4

We add all the numbers together, and all the variables

2b+2

Back to the equation:

-(2b+2)



We get rid of parentheses

-2b-2+3

We add all the numbers together, and all the variables

-2b+1

Back to the equation:

-(-2b+1)



We add all the numbers together, and all the variables

-1b-(-2b+1)-6=0

We get rid of parentheses

-1b+2b-1-6=0

We add all the numbers together, and all the variables

b-7=0

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

b=7

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