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

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



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

We move all terms to the left:

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



Domain of the equation: 6(x+2)!=0

x∈R





Domain of the equation: 3(4-x))!=0

x∈R



We add all the numbers together, and all the variables

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

We calculate fractions


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



We calculate terms in parentheses: +(3x(-)/(6(x+2)*3(-1x+4))), so:

3x(-)/(6(x+2)*3(-1x+4))

We add all the numbers together, and all the variables

3x0/(6(x+2)*3(-1x+4))

We multiply all the terms by the denominator


3x0

We add all the numbers together, and all the variables

3x

Back to the equation:

+(3x)





We calculate terms in parentheses: +(-6xx/(6(x+2)*3(-1x+4))), so:

-6xx/(6(x+2)*3(-1x+4))

We multiply all the terms by the denominator


-6xx

Back to the equation:

+(-6xx)



We get rid of parentheses

3x-6xx=0

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