b=16+2(4-b)+3

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Solution for b=16+2(4-b)+3 equation: 



b=16+2(4-b)+3

We move all terms to the left:

b-(16+2(4-b)+3)=0

We add all the numbers together, and all the variables

b-(16+2(-1b+4)+3)=0



We calculate terms in parentheses: -(16+2(-1b+4)+3), so:

16+2(-1b+4)+3

determiningTheFunctionDomain
2(-1b+4)+16+3

We add all the numbers together, and all the variables

2(-1b+4)+19

We multiply parentheses

-2b+8+19

We add all the numbers together, and all the variables

-2b+27

Back to the equation:

-(-2b+27)



We get rid of parentheses

b+2b-27=0

We add all the numbers together, and all the variables

3b-27=0

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

3b=27

b=27/3

b=9

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