(7x-3)/3=(3x-4)/8

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



(7x-3)/3=(3x-4)/8

We move all terms to the left:

(7x-3)/3-((3x-4)/8)=0

We calculate fractions


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



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

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

We multiply all the terms by the denominator


-((3x-4)*3)



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

(3x-4)*3

We multiply parentheses

9x-12

Back to the equation:

-(9x-12)



We get rid of parentheses

-9x+12

Back to the equation:

+(-9x+12)



We get rid of parentheses

7x/()-9x+12=0

We multiply all the terms by the denominator


7x-9x*()+12*()=0

We add all the numbers together, and all the variables

7x-9x*()=0

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