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

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



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

We move all terms to the left:

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



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

k∈R





Domain of the equation: 5(k+1))!=0

k∈R



We calculate fractions


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



We calculate terms in parentheses: +(5kk/(4(k-3)*5(k+1))), so:

5kk/(4(k-3)*5(k+1))

We multiply all the terms by the denominator


5kk

Back to the equation:

+(5kk)





We calculate terms in parentheses: +(-4kk/(4(k-3)*5(k+1))), so:

-4kk/(4(k-3)*5(k+1))

We multiply all the terms by the denominator


-4kk

Back to the equation:

+(-4kk)



We get rid of parentheses

5kk-4kk=0

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