(3x-4)/9+x=(3x+4)/3+1

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Solution for (3x-4)/9+x=(3x+4)/3+1 equation: 



(3x-4)/9+x=(3x+4)/3+1

We move all terms to the left:

(3x-4)/9+x-((3x+4)/3+1)=0

We add all the numbers together, and all the variables

x+(3x-4)/9-((3x+4)/3+1)=0

We calculate fractions


x+(9x-12)/()+(-((3x+4)*9)/()=0



We calculate terms in parentheses: +(-((3x+4)*9)/(), so:

-((3x+4)*9)/(

We multiply all the terms by the denominator


-((3x+4)*9)



We calculate terms in parentheses: -((3x+4)*9), so:

(3x+4)*9

We multiply parentheses

27x+36

Back to the equation:

-(27x+36)



We get rid of parentheses

-27x-36

Back to the equation:

+(-27x-36)



We get rid of parentheses

x+(9x-12)/()-27x-36=0

We multiply all the terms by the denominator


x*()+(9x-12)-27x*()-36*()=0

We add all the numbers together, and all the variables

x*()+(9x-12)-27x*()=0

We get rid of parentheses

x*()+9x-27x*()-12=0

We add all the numbers together, and all the variables

9x+x*()-27x*()-12=0

We move all terms containing x to the left, all other terms to the right

9x+x*()-27x*()=12

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