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?
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