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

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



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

We move all terms to the left:

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



Domain of the equation: 2b+(b+45)+(2b-90)+90)!=0

b∈R



We multiply all the terms by the denominator


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



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

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

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

We add all the numbers together, and all the variables

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

Wy multiply elements

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

We get rid of parentheses

b+1080b+b+2b+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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