(2x+1)/8=(3x-4)/7

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Solution for (2x+1)/8=(3x-4)/7 equation: 



(2x+1)/8=(3x-4)/7

We move all terms to the left:

(2x+1)/8-((3x-4)/7)=0

We calculate fractions


2x/()+(-((3x-4)*8)/()=0



We calculate terms in parentheses: +(-((3x-4)*8)/(), so:

-((3x-4)*8)/(

We multiply all the terms by the denominator


-((3x-4)*8)



We calculate terms in parentheses: -((3x-4)*8), so:

(3x-4)*8

We multiply parentheses

24x-32

Back to the equation:

-(24x-32)



We get rid of parentheses

-24x+32

Back to the equation:

+(-24x+32)



We get rid of parentheses

2x/()-24x+32=0

We multiply all the terms by the denominator


2x-24x*()+32*()=0

We add all the numbers together, and all the variables

2x-24x*()=0

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