(2x+1)/7=(3x-4)/6

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Solution for (2x+1)/7=(3x-4)/6 equation: 



(2x+1)/7=(3x-4)/6

We move all terms to the left:

(2x+1)/7-((3x-4)/6)=0

We calculate fractions


2x/()+(-((3x-4)*7)/()=0



We calculate terms in parentheses: +(-((3x-4)*7)/(), so:

-((3x-4)*7)/(

We multiply all the terms by the denominator


-((3x-4)*7)



We calculate terms in parentheses: -((3x-4)*7), so:

(3x-4)*7

We multiply parentheses

21x-28

Back to the equation:

-(21x-28)



We get rid of parentheses

-21x+28

Back to the equation:

+(-21x+28)



We get rid of parentheses

2x/()-21x+28=0

We multiply all the terms by the denominator


2x-21x*()+28*()=0

We add all the numbers together, and all the variables

2x-21x*()=0

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