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

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Solution for 9(x-3)/6=3(x-+2)/7 equation: 



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

We move all terms to the left:

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

We add all the numbers together, and all the variables

9(x-3)/6-(3(+x)/7)=0

We calculate fractions


9x/()+(-(3(+x)*6)/()=0



We calculate terms in parentheses: +(-(3(+x)*6)/(), so:

-(3(+x)*6)/(

We multiply all the terms by the denominator


-(3(+x)*6)



We calculate terms in parentheses: -(3(+x)*6), so:

3(+x)*6

We multiply parentheses

18x

Back to the equation:

-(18x)



Back to the equation:

+(-18x)



We get rid of parentheses

9x/()-18x=0

We multiply all the terms by the denominator


9x-18x*()=0

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