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

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Solution for (2/3)k-(k+(1/3))=(1/9)(k+3) equation: 



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

We move all terms to the left:

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



Domain of the equation: 3)k!=0

k!=0/1

k!=0

k∈R





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

k∈R



We add all the numbers together, and all the variables

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

We multiply parentheses

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

We multiply parentheses ..

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

We calculate fractions


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

We multiply all the terms by the denominator


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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-1k^2+2k^2*()-1k-30=0

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

-1k^2+2k^2*()-1k=30

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