A=1/35(1+4b+1)

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Solution for A=1/35(1+4b+1) equation: 



=1/35(1+4A+1)

We move all terms to the left:

-(1/35(1+4A+1))=0



Domain of the equation: 35(1+4A+1))!=0

A∈R



We add all the numbers together, and all the variables

-(1/35(4A+2))=0

We multiply all the terms by the denominator


-(1=0

We add all the numbers together, and all the variables

=0

A=0/1

A=0

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