1/3k-(k+1-3)=1/9(k+3)

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Solution for 1/3k-(k+1-3)=1/9(k+3) equation: 



1/3k-(k+1-3)=1/9(k+3)

We move all terms to the left:

1/3k-(k+1-3)-(1/9(k+3))=0



Domain of the equation: 3k!=0

k!=0/3

k!=0

k∈R





Domain of the equation: 9(k+3))!=0

k∈R



We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We calculate fractions


-k+(9kk/(27k^2k+(-(1*3k)/(27k^2k+2=0

We can not solve this equation

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