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

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



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

We move all terms to the left:

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

We calculate fractions


3x/()+(-((2x+2)*9)/()=0



We calculate terms in parentheses: +(-((2x+2)*9)/(), so:

-((2x+2)*9)/(

We multiply all the terms by the denominator


-((2x+2)*9)



We calculate terms in parentheses: -((2x+2)*9), so:

(2x+2)*9

We multiply parentheses

18x+18

Back to the equation:

-(18x+18)



We get rid of parentheses

-18x-18

Back to the equation:

+(-18x-18)



We get rid of parentheses

3x/()-18x-18=0

We multiply all the terms by the denominator


3x-18x*()-18*()=0

We add all the numbers together, and all the variables

3x-18x*()=0

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