Hewo!

WHAT I HAVE SO FAR:

J+K           1

___     =  _____

JK              3

Right?

What can I do from here?

I have not worked out how to do it formally and I won't guarentee that what i think is correct.

I don't think what you have done is helpful because the answers for j and k must be integers

I have only found 1 solution set for j and k, well 2 really because k and j are interchangable.

Can you find one answer, just by mucking around with the numbers?

Not sure if it helps but,

Solving for k

k= 3j/j-3  , Notice J can't be equal to 3 

If J and K are positive integers, 

Is there a restricted domain?

Also J can't be equal to=(1,2,3,5,7,8,9,10,11,13,14,15,16,17,18,19,20,21,22, etc..) The pattern contuines so, k can be equal to 4, 6 ,12only out of the first 20 numbers.. I don't think the equation works for more... Idk. 

so the sum of possible values of k is 4+6+12=22 

https://www.desmos.com/calculator 

Type your question here but let k=x and j=y or vice versa and notice that k=j at 6 

also when k or j = 4 the other = 12 

Also notice that it will approach 3 but can't equal to 3 as it will be undefined. 

Here is my attempt:

Good job on finding \( 0=-3j+jk-3k\) guys, we can work from here?

We factor:

\(0=j(-3+k)-3k\)

Simplifies to:

\(0=j(k-3)-3k\)

We add 9 to both sides, this allows us to factor better:

\(9=j(k-3)-3k+9\)

We factor once again

\(9=j(k-3)-3(k-3)\)

We then simplify:

\(9=(j-3)(k-3)\)

Now that we have a binomial expression, we can list the factors of 9 to guess solutions.

1 and 9

3 and 3.

So for 1 and 9, we set them equal

\(j-3=1\)

\(k-3=9\)

Solving this, we get J=4 and k=12, since they can be reversed, the values of K and be 4 or 12.

Now we solve 3 and 3

\(j-3=3\)

\(k-3=3\)

Solving this, we get J=6 and K=6, since they can be reversed, the values of K can only be 6.

So the possible values for K can be 4, 6, and 12.

The sum of that is \(\boxed{22}\)

A factor of 9 is also -3 and -3. But do you know why it doesn't work?