540=b+3/2b+(b+45)

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Solution for 540=b+3/2b+(b+45) equation: 



540=b+3/2b+(b+45)

We move all terms to the left:

540-(b+3/2b+(b+45))=0



Domain of the equation: 2b+(b+45))!=0

b∈R



We multiply all the terms by the denominator


-(b+3+540*2b+(b+45))=0



We calculate terms in parentheses: -(b+3+540*2b+(b+45)), so:

b+3+540*2b+(b+45)

determiningTheFunctionDomain
b+540*2b+(b+45)+3

Wy multiply elements

b+1080b+(b+45)+3

We get rid of parentheses

b+1080b+b+45+3

We add all the numbers together, and all the variables

1082b+48

Back to the equation:

-(1082b+48)



We get rid of parentheses

-1082b-48=0

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

-1082b=48

b=48/-1082

b=-24/541

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