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1/2(k+3)=1/3(k+16)
We move all terms to the left:
1/2(k+3)-(1/3(k+16))=0
Domain of the equation: 2(k+3)!=0
k∈R
Domain of the equation: 3(k+16))!=0We calculate fractions
k∈R
(3kk/(2(k+3)*3(k+16)))+(-2kk/(2(k+3)*3(k+16)))=0
We calculate terms in parentheses: +(3kk/(2(k+3)*3(k+16))), so:
3kk/(2(k+3)*3(k+16))
We multiply all the terms by the denominator
3kk
Back to the equation:
+(3kk)
We calculate terms in parentheses: +(-2kk/(2(k+3)*3(k+16))), so:We get rid of parentheses
-2kk/(2(k+3)*3(k+16))
We multiply all the terms by the denominator
-2kk
Back to the equation:
+(-2kk)
3kk-2kk=0
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