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In the equation $\frac{1}{j} + \frac{1}{k} = \frac{1}{3}$, both $j$ and $k$ are positive integers. What is the sum of all possible values for $k$?


Sagot :

Answer:

ayo ion get the equation. type up it again?

Step-by-step explanation:

caylus

Answer:

Hello,

answer 22

Step-by-step explanation:

[tex]\dfrac{1}{j} +\dfrac{1}{k} =\dfrac{1}{3} \\\\\dfrac{k+j}{j*k} =\dfrac{1}{3} \\\\\\3k+3j=j*k\\\\k(3-j)=-3j\\\\k=\dfrac{3j}{j-3} \\\\k=\dfrac{3j-9+9}{j-3} \\\\k=3+\dfrac{9}{j-3} \\\\\\j-3\ must\ be \ a divisor\ of\ 9 ==> 1,3,9\\\\j-3=1 ==> j=4 , k=3+\dfrac{9}{1} =12\\\\j-3=3 ==> j=6 , k=3+\dfrac{9}{3} =6\\\\j-3=9 ==> j=12 , k=3+\dfrac{9}{9} =4\\\\\sum\ k=12+6+4=22\\[/tex]