(3x+1)/7=(2x-5)/6

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Solution for (3x+1)/7=(2x-5)/6 equation: 



(3x+1)/7=(2x-5)/6

We move all terms to the left:

(3x+1)/7-((2x-5)/6)=0

We calculate fractions


3x/()+(-((2x-5)*7)/()=0



We calculate terms in parentheses: +(-((2x-5)*7)/(), so:

-((2x-5)*7)/(

We multiply all the terms by the denominator


-((2x-5)*7)



We calculate terms in parentheses: -((2x-5)*7), so:

(2x-5)*7

We multiply parentheses

14x-35

Back to the equation:

-(14x-35)



We get rid of parentheses

-14x+35

Back to the equation:

+(-14x+35)



We get rid of parentheses

3x/()-14x+35=0

We multiply all the terms by the denominator


3x-14x*()+35*()=0

We add all the numbers together, and all the variables

3x-14x*()=0

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