(2x-3)/6=(3x+1)/2

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Solution for (2x-3)/6=(3x+1)/2 equation: 



(2x-3)/6=(3x+1)/2

We move all terms to the left:

(2x-3)/6-((3x+1)/2)=0

We calculate fractions


2x/()+(-((3x+1)*6)/()=0



We calculate terms in parentheses: +(-((3x+1)*6)/(), so:

-((3x+1)*6)/(

We multiply all the terms by the denominator


-((3x+1)*6)



We calculate terms in parentheses: -((3x+1)*6), so:

(3x+1)*6

We multiply parentheses

18x+6

Back to the equation:

-(18x+6)



We get rid of parentheses

-18x-6

Back to the equation:

+(-18x-6)



We get rid of parentheses

2x/()-18x-6=0

We multiply all the terms by the denominator


2x-18x*()-6*()=0

We add all the numbers together, and all the variables

2x-18x*()=0

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