1/4(2k-6)=1/12(3k+12)1

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Solution for 1/4(2k-6)=1/12(3k+12)1 equation: 



1/4(2k-6)=1/12(3k+12)1

We move all terms to the left:

1/4(2k-6)-(1/12(3k+12)1)=0



Domain of the equation: 4(2k-6)!=0

k∈R





Domain of the equation: 12(3k+12)1)!=0

k∈R



We calculate fractions


(12k3/(4(2k-6)*12(3k+12)1))+(-4k2/(4(2k-6)*12(3k+12)1))=0



We calculate terms in parentheses: +(12k3/(4(2k-6)*12(3k+12)1)), so:

12k3/(4(2k-6)*12(3k+12)1)

We multiply all the terms by the denominator


12k3

We add all the numbers together, and all the variables

12k^3

We do not support ekpression: k^3

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