(5x)/7=(2x-4)/3

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Solution for (5x)/7=(2x-4)/3 equation: 



(5x)/7=(2x-4)/3

We move all terms to the left:

(5x)/7-((2x-4)/3)=0

We calculate fractions


15x^2/()+(-((2x-4)*7)/()=0



We calculate terms in parentheses: +(-((2x-4)*7)/(), so:

-((2x-4)*7)/(

We multiply all the terms by the denominator


-((2x-4)*7)



We calculate terms in parentheses: -((2x-4)*7), so:

(2x-4)*7

We multiply parentheses

14x-28

Back to the equation:

-(14x-28)



We get rid of parentheses

-14x+28

Back to the equation:

+(-14x+28)



We get rid of parentheses

15x^2/()-14x+28=0

We multiply all the terms by the denominator


15x^2-14x*()+28*()=0

We add all the numbers together, and all the variables

15x^2-14x*()=0

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