(3x+1)/3-(13/2)=(1-x)/3

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Solution for (3x+1)/3-(13/2)=(1-x)/3 equation: 



(3x+1)/3-(13/2)=(1-x)/3

We move all terms to the left:

(3x+1)/3-(13/2)-((1-x)/3)=0

We add all the numbers together, and all the variables

(3x+1)/3-((-1x+1)/3)-(+13/2)=0

We get rid of parentheses

(3x+1)/3-((-1x+1)/3)-13/2=0

We calculate fractions


(6x+2)/()+(-((-1x+1)*2)/()+()/()=0



We calculate terms in parentheses: +(-((-1x+1)*2)/()+()/(), so:

-((-1x+1)*2)/()+()/(

We add all the numbers together, and all the variables

-((-1x+1)*2)/()+1

We multiply all the terms by the denominator


-((-1x+1)*2)+1*()



We calculate terms in parentheses: -((-1x+1)*2), so:

(-1x+1)*2

We multiply parentheses

-2x+2

Back to the equation:

-(-2x+2)



We add all the numbers together, and all the variables

-(-2x+2)

We get rid of parentheses

2x-2

Back to the equation:

+(2x-2)



We get rid of parentheses

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

We multiply all the terms by the denominator


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

We add all the numbers together, and all the variables

(6x+2)+2x*()=0

We get rid of parentheses

6x+2x*()+2=0

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

6x+2x*()=-2

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