(k+3)/2k=(k-8)/3k

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Solution for (k+3)/2k=(k-8)/3k equation: 



(k+3)/2k=(k-8)/3k

We move all terms to the left:

(k+3)/2k-((k-8)/3k)=0



Domain of the equation: 2k!=0

k!=0/2

k!=0

k∈R





Domain of the equation: 3k)!=0

k!=0/1

k!=0

k∈R



We calculate fractions


(k+3)*3k)/6k^2+(-((k-8)*2k)/6k^2=0

We calculate fractions


((k+3)*3k)*6k^2)/(6k^2+(*6k^2)+(-((k-8)*2k)*6k^2)/(6k^2+(*6k^2)=0



We calculate terms in parentheses: +(-((k-8)*2k)*6k^2)/(6k^2+(*6k^2), so:

-((k-8)*2k)*6k^2)/(6k^2+(*6k^2

We multiply all the terms by the denominator


-((k-8)*2k)*6k^2)+((*6k^2)*(6k^2

Back to the equation:

+(-((k-8)*2k)*6k^2)+((*6k^2)*(6k^2)



We get rid of parentheses

((k+3)*3k)*6k^2)/(6k^2+*6k^2+(-((k-8)*2k)*6k^2)+((*6k^2)*6k^2=0

We multiply all the terms by the denominator


((k+3)*3k)*6k^2)+(*6k^2)*(6k^2+((-((k-8)*2k)*6k^2))*(6k^2+(((*6k^2)*6k^2)*(6k^2=0

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