(3b-3)/2=(b-4)/6

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Solution for (3b-3)/2=(b-4)/6 equation: 



(3b-3)/2=(b-4)/6

We move all terms to the left:

(3b-3)/2-((b-4)/6)=0

We calculate fractions


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



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

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

We multiply all the terms by the denominator


-((b-4)*2)



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

(b-4)*2

We multiply parentheses

2b-8

Back to the equation:

-(2b-8)



We get rid of parentheses

-2b+8

Back to the equation:

+(-2b+8)



We get rid of parentheses

3b/()-2b+8=0

We multiply all the terms by the denominator


3b-2b*()+8*()=0

We add all the numbers together, and all the variables

3b-2b*()=0

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