(6x+2)/3+4/3=2x+1+(x+1)/6

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Solution for (6x+2)/3+4/3=2x+1+(x+1)/6 equation: 



(6x+2)/3+4/3=2x+1+(x+1)/6

We move all terms to the left:

(6x+2)/3+4/3-(2x+1+(x+1)/6)=0

We calculate fractions


(6x+2)/()+(-(2x+1+(x+1)*3)/()=0



We calculate terms in parentheses: +(-(2x+1+(x+1)*3)/(), so:

-(2x+1+(x+1)*3)/(

We multiply all the terms by the denominator


-(2x+1+(x+1)*3)



We calculate terms in parentheses: -(2x+1+(x+1)*3), so:

2x+1+(x+1)*3

determiningTheFunctionDomain
2x+(x+1)*3+1

We multiply parentheses

2x+3x+3+1

We add all the numbers together, and all the variables

5x+4

Back to the equation:

-(5x+4)



We get rid of parentheses

-5x-4

Back to the equation:

+(-5x-4)



We get rid of parentheses

(6x+2)/()-5x-4=0

We multiply all the terms by the denominator


(6x+2)-5x*()-4*()=0

We add all the numbers together, and all the variables

(6x+2)-5x*()=0

We get rid of parentheses

6x-5x*()+2=0

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

6x-5x*()=-2

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