16=(1/2)(4)(3+b1)

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Solution for 16=(1/2)(4)(3+b1) equation: 



16=(1/2)(4)(3+b1)

We move all terms to the left:

16-((1/2)(4)(3+b1))=0



Domain of the equation: 2)4(3+b1))!=0

b∈R



We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


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



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

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

We add all the numbers together, and all the variables

334(b+3)

We multiply parentheses

334b+1002

Back to the equation:

-(334b+1002)



We get rid of parentheses

-334b-1002=0

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

-334b=1002

b=1002/-334

b=-3

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