The idea is to form the same coefficient for either s or t in both equations. So if we want to eliminate s, I guess the easiest way is to find the common multiple, which would be 15, so:
5s - 3t - 13 = 0 (each term x3)
3s - 7t - 13 = 0 (each term x5)
15s - 9t - 39 = 0
15s - 35t - 65 = 0
So subtracting the second equation from the first gives:
26t + 26 = 0
(It may be easier to label your equations, maybe with (1) and (2) for the first and second equations, then write: (1) - (2) to show you are subtracting the two equations)
(Also note, the 15s and 15s cancel each other out, which is idea of the elimination method. This leaves just one unknown variable)
Clearly t = -1, then you can sub that into either of the original equations which gives s = 2.
edit: too slow