(5(2k-3)-3(k+4)/3k+2)k=2

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Solution for (5(2k-3)-3(k+4)/3k+2)k=2 equation: 



(5(2k-3)-3(k+4)/3k+2)k=2

We move all terms to the left:

(5(2k-3)-3(k+4)/3k+2)k-(2)=0



Domain of the equation: 3k+2)k!=0

k∈R



We multiply all the terms by the denominator


(5(2k-3)-3(k+4)-2*3k+2)k=0

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