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

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



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

We move all terms to the left:

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

We calculate fractions


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



We calculate terms in parentheses: +(-(2(x+1)*6)/(), so:

-(2(x+1)*6)/(

We multiply all the terms by the denominator


-(2(x+1)*6)



We calculate terms in parentheses: -(2(x+1)*6), so:

2(x+1)*6

We multiply parentheses

12x+12

Back to the equation:

-(12x+12)



We get rid of parentheses

-12x-12

Back to the equation:

+(-12x-12)



We get rid of parentheses

4x/()-12x-12=0

We multiply all the terms by the denominator


4x-12x*()-12*()=0

We add all the numbers together, and all the variables

4x-12x*()=0

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