(4(b+2))/3=(3b-49)/6

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Solution for (4(b+2))/3=(3b-49)/6 equation: 



(4(b+2))/3=(3b-49)/6

We move all terms to the left:

(4(b+2))/3-((3b-49)/6)=0

We calculate fractions


24b/()+(-((3b-49)*3)/()=0



We calculate terms in parentheses: +(-((3b-49)*3)/(), so:

-((3b-49)*3)/(

We multiply all the terms by the denominator


-((3b-49)*3)



We calculate terms in parentheses: -((3b-49)*3), so:

(3b-49)*3

We multiply parentheses

9b-147

Back to the equation:

-(9b-147)



We get rid of parentheses

-9b+147

Back to the equation:

+(-9b+147)



We get rid of parentheses

24b/()-9b+147=0

We multiply all the terms by the denominator


24b-9b*()+147*()=0

We add all the numbers together, and all the variables

24b-9b*()=0

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