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

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



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

We move all terms to the left:

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



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

a∈R





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

a∈R



We add all the numbers together, and all the variables

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

We calculate fractions


(3a(-)/(6(a+2)*3(-1a+4)))+(-6aa/(6(a+2)*3(-1a+4)))=0



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

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

We add all the numbers together, and all the variables

3a0/(6(a+2)*3(-1a+4))

We multiply all the terms by the denominator


3a0

We add all the numbers together, and all the variables

3a

Back to the equation:

+(3a)





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

-6aa/(6(a+2)*3(-1a+4))

We multiply all the terms by the denominator


-6aa

Back to the equation:

+(-6aa)



We get rid of parentheses

3a-6aa=0

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