1/3k-3+1=1/3(k-6)=

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Solution for 1/3k-3+1=1/3(k-6)= equation: 



1/3k-3+1=1/3(k-6)=

We move all terms to the left:

1/3k-3+1-(1/3(k-6))=0



Domain of the equation: 3k!=0

k!=0/3

k!=0

k∈R





Domain of the equation: 3(k-6))!=0

k∈R



We add all the numbers together, and all the variables

1/3k-(1/3(k-6))-2=0

We calculate fractions


(3kk/(9k^2k+(-(1*3k)/(9k^2k-2=0

We can not solve this equation

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