Algebra

The only way for the system to have infinitely solutions is where we have two exactly congruent equations. 

We first notice that we have

$6a + 2b = k + 3a + mb$

Subtracting 3a from both sides will achieve us the same exact left sides of the equations. We have

$3a+2b=2\\ 3a+2b = k+mb$

Setting the right sides of the equation to equal each other, we have

$k+mb=2$

Now, let's note that if m was any number other than 0, the two equations would fail to match each other. Therefore, m must be 0. 

For this value of m, we have k is 2. So we have

$m=0\\ k=2$

So our final answer is m = 0 and k = 2. 

Thanks! :)