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

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



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

We move all terms to the left:

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



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

k∈R





Domain of the equation: 32k!=0

k!=0/32

k!=0

k∈R



We calculate fractions


96k/(64k^2-192k)+(-4kk/(64k^2-192k)=0



We calculate terms in parentheses: +(-4kk/(64k^2-192k), so:

-4kk/(64k^2-192k

We multiply all the terms by the denominator


-4kk

Back to the equation:

+(-4kk)



We get rid of parentheses

96k/(64k^2-192k)-4kk=0

We multiply all the terms by the denominator


96k-4kk*(64k^2-192k)=0

We multiply parentheses

-256k^3+768k^2+96k=0

We do not support ekpression: k^3

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