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

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

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

We move all terms to the left:

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


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

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

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

b∈R


We multiply all the terms by the denominator


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


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

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

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

We add all the numbers together, and all the variables

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

Wy multiply elements

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

1084b+228

Back to the equation:

-(1084b+228)


We get rid of parentheses

-1084b-228=0

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

-1084b=228

b=228/-1084

b=-57/271