For this system:

I don't think I can multiply through by

because those aren't necessarily constants.

Mr F says: The unknowns you're solving for are x and y. So yes, you can.
And therefore that goes against the elementary rules I need to follow for this assignment at least. So, in the end I would up subtracting the first equation from the second one (add -1 * first equation to second one) and then solved for y in the second one. Then proceeded to plug the answer for y into the first one. Did you get the solution I got for that system?

Also, my work for the second set:

1. kx + y = 4

2. 2x - 3y = -12

Went like this:

Multiply equation 1 by 3 and replace equation 1 with the result (3*eq1 ->eq1) . Then add equation 1 to equation 2 and replace equation 2 with the result. So the system will then look like this:

1. 3kx + 3y = 12

2. (3k + 2)x = 0

Equation 2 is true when either x = 0 or (3k + 2) = 0. Since x = 0 is trivial I solved 3k + 2 = 0 for k and got k = -2/3.

With that we can put in any value for x and there will be a corresponding value for y that will make the system true. For example (3,6) or (6, 8). Since we know a system can have only 1 unique solution, no solution, or infinitely many solutions, the fact that we can find two points which satisfy the system when k = -2/3 means it has an infinite number of solutions. I guess that ought to be good enough, There wouldn't be another value for k to make it true.

Mr F says: I guess so.