(5)/(3+b)=(3)/(7b=1)

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Solution for (5)/(3+b)=(3)/(7b=1) equation: 



(5)/(3+b)=(3)/(7b=1)

We move all terms to the left:

(5)/(3+b)-((3)/(7b)=0



Domain of the equation: (3+b)!=0

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

b!=-3

b∈R





Domain of the equation: 7b!=0

b!=0/7

b!=0

b∈R



We add all the numbers together, and all the variables

5/(b+3)-(3/7b=0

We calculate fractions


35b/(7b^2+21b)+(-(3*(b+3))/(7b^2+21b)=0



We calculate terms in parentheses: +(-(3*(b+3))/(7b^2+21b), so:

-(3*(b+3))/(7b^2+21b

We multiply all the terms by the denominator


-(3*(b+3))



We calculate terms in parentheses: -(3*(b+3)), so:

3*(b+3)

We multiply parentheses

3b+9

Back to the equation:

-(3b+9)



We get rid of parentheses

-3b-9

Back to the equation:

+(-3b-9)



We get rid of parentheses

35b/(7b^2+21b)-3b-9=0

We multiply all the terms by the denominator


35b-3b*(7b^2+21b)-9*(7b^2+21b)=0

We multiply parentheses

-21b^3-63b^2-63b^2+35b-189b=0

We do not support ebpression: b^3

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