(n+2)/3=(6-n)/9

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



(n+2)/3=(6-n)/9

We move all terms to the left:

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

We add all the numbers together, and all the variables

(n+2)/3-((-1n+6)/9)=0

We calculate fractions


n/()+(-((-1n+6)*3)/()=0



We calculate terms in parentheses: +(-((-1n+6)*3)/(), so:

-((-1n+6)*3)/(

We multiply all the terms by the denominator


-((-1n+6)*3)



We calculate terms in parentheses: -((-1n+6)*3), so:

(-1n+6)*3

We multiply parentheses

-3n+18

Back to the equation:

-(-3n+18)



We get rid of parentheses

3n-18

Back to the equation:

+(3n-18)



We get rid of parentheses

n/()+3n-18=0

We multiply all the terms by the denominator


n+3n*()-18*()=0

We add all the numbers together, and all the variables

n+3n*()=0

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