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

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



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

We move all terms to the left:

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

We calculate fractions


6x/()+(-(2(x+2)*3)/()=0



We calculate terms in parentheses: +(-(2(x+2)*3)/(), so:

-(2(x+2)*3)/(

We multiply all the terms by the denominator


-(2(x+2)*3)



We calculate terms in parentheses: -(2(x+2)*3), so:

2(x+2)*3

We multiply parentheses

6x+12

Back to the equation:

-(6x+12)



We get rid of parentheses

-6x-12

Back to the equation:

+(-6x-12)



We get rid of parentheses

6x/()-6x-12=0

We multiply all the terms by the denominator


6x-6x*()-12*()=0

We add all the numbers together, and all the variables

6x-6x*()=0

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