5/12(k-2)+1/4(3k-4)=k+7/2

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Solution for 5/12(k-2)+1/4(3k-4)=k+7/2 equation: 



5/12(k-2)+1/4(3k-4)=k+7/2

We move all terms to the left:

5/12(k-2)+1/4(3k-4)-(k+7/2)=0



Domain of the equation: 12(k-2)!=0

k∈R





Domain of the equation: 4(3k-4)!=0

k∈R



We add all the numbers together, and all the variables

5/12(k-2)+1/4(3k-4)-(+k+7/2)=0

We get rid of parentheses

5/12(k-2)+1/4(3k-4)-k-7/2=0

We calculate fractions


(40k3/(12(k-2)*4(3k-4)*2)+(24kk/(12(k-2)*4(3k-4)*2)+(-336k^2k/(12(k-2)*4(3k-4)*2)-k=0

We can not solve this equation

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