540=b+(b+45)+90+(2b-90)+3/2b

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Solution for 540=b+(b+45)+90+(2b-90)+3/2b equation: 



540=b+(b+45)+90+(2b-90)+3/2b

We move all terms to the left:

540-(b+(b+45)+90+(2b-90)+3/2b)=0



Domain of the equation: 2b)!=0

b!=0/1

b!=0

b∈R



We multiply all the terms by the denominator


-(b+(b+45)+90+(2b-90)+3+540*2b)=0



We calculate terms in parentheses: -(b+(b+45)+90+(2b-90)+3+540*2b), so:

b+(b+45)+90+(2b-90)+3+540*2b

determiningTheFunctionDomain
b+(b+45)+(2b-90)+540*2b+90+3

We add all the numbers together, and all the variables

b+(b+45)+(2b-90)+540*2b+93

Wy multiply elements

b+(b+45)+(2b-90)+1080b+93

We get rid of parentheses

b+b+2b+1080b+45-90+93

We add all the numbers together, and all the variables

1084b+48

Back to the equation:

-(1084b+48)



We get rid of parentheses

-1084b-48=0

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

-1084b=48

b=48/-1084

b=-12/271

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