15=4+2(b+3)

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Solution for 15=4+2(b+3) equation: 



15=4+2(b+3)

We move all terms to the left:

15-(4+2(b+3))=0



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

4+2(b+3)

determiningTheFunctionDomain
2(b+3)+4

We multiply parentheses

2b+6+4

We add all the numbers together, and all the variables

2b+10

Back to the equation:

-(2b+10)



We get rid of parentheses

-2b-10+15=0

We add all the numbers together, and all the variables

-2b+5=0

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

-2b=-5

b=-5/-2

b=2+1/2

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