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540=b+3/2b+(b+45)+(2b-90)+90+b
We move all terms to the left:
540-(b+3/2b+(b+45)+(2b-90)+90+b)=0
Domain of the equation: 2b+(b+45)+(2b-90)+90+b)!=0We multiply all the terms by the denominator
We move all terms containing b to the left, all other terms to the right
2b+(b+45)+(2b-90)+b)!=-90
b∈R
-(b+3+540*2b+(b+45)+(2b-90)+90+b)=0
We calculate terms in parentheses: -(b+3+540*2b+(b+45)+(2b-90)+90+b), so:We get rid of parentheses
b+3+540*2b+(b+45)+(2b-90)+90+b
determiningTheFunctionDomain b+540*2b+(b+45)+(2b-90)+b+3+90
We add all the numbers together, and all the variables
2b+540*2b+(b+45)+(2b-90)+93
Wy multiply elements
2b+1080b+(b+45)+(2b-90)+93
We get rid of parentheses
2b+1080b+b+2b+45-90+93
We add all the numbers together, and all the variables
1085b+48
Back to the equation:
-(1085b+48)
-1085b-48=0
We move all terms containing b to the left, all other terms to the right
-1085b=48
b=48/-1085
b=-48/1085
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