3/(3b+4)=2/(b-4)

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Solution for 3/(3b+4)=2/(b-4) equation: 



3/(3b+4)=2/(b-4)

We move all terms to the left:

3/(3b+4)-(2/(b-4))=0



Domain of the equation: (3b+4)!=0

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

3b!=-4

b!=-4/3

b!=-1+1/3

b∈R





Domain of the equation: (b-4))!=0

b∈R



We calculate fractions


3b/((3b+4)*(b-4)))+(-(2*(3b+4))/((3b+4)*(b-4)))=0



We calculate terms in parentheses: -(2*(3b+4))/((3b+4)*(b-4))), so:

2*(3b+4))/((3b+4)*(b-4))

We multiply all the terms by the denominator


2*(3b+4))

We multiply parentheses

6b+

We add all the numbers together, and all the variables

6b

Back to the equation:

-(6b)



We add all the numbers together, and all the variables

-6b+3b/((3b+4)*(b-4)))+(=0

We multiply all the terms by the denominator


-6b*((3b+4)*(b-4)))+(+3b=0

We add all the numbers together, and all the variables

3b-6b*((3b+4)*(b-4)))+(=0

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