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

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

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

We move all terms to the left:

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


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

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

2b+b+(b+45)+(2b-90))!=-90

b∈R


We multiply all the terms by the denominator


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


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

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

determiningTheFunctionDomain
540*2b+b+(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