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

Solution for 4/(b+3)+5/b=1 equation:

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

We move all terms to the left:

4/(b+3)+5/b-(1)=0


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

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

b!=-3

b∈R



Domain of the equation: b!=0

b∈R


We calculate fractions


4b/(b^2+3b)+(5b+15)/(b^2+3b)-1=0

We multiply all the terms by the denominator


4b+(5b+15)-1*(b^2+3b)=0

We get rid of parentheses

4b+5b-1*(b^2+3b)+15=0

We add all the numbers together, and all the variables

9b-1*(b^2+3b)+15=0

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

9b-1*(b^2+3b)=-15