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

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



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

We move all terms to the left:

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



Domain of the equation: 5(k+2)!=0

k∈R





Domain of the equation: 5(k-4))!=0

k∈R



We calculate fractions


(5kk/(5(k+2)*5(k-4)))+(-15kk/(5(k+2)*5(k-4)))=0



We calculate terms in parentheses: +(5kk/(5(k+2)*5(k-4))), so:

5kk/(5(k+2)*5(k-4))

We multiply all the terms by the denominator


5kk

Back to the equation:

+(5kk)





We calculate terms in parentheses: +(-15kk/(5(k+2)*5(k-4))), so:

-15kk/(5(k+2)*5(k-4))

We multiply all the terms by the denominator


-15kk

Back to the equation:

+(-15kk)



We get rid of parentheses

5kk-15kk=0

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