Which constants can be multiplied by the equations so one variable will be eliminated when the systems are added together?
5x + 13y = 232
12x + 7y = 218
the first equation can be multiplied by –13 and the second equation by 7 to eliminate y.
the first equation can be multiplied by 7 and the second equation by 13 to eliminate y.
the first equation can be multiplied by –12 and the second equation by 5 to eliminate x.
the first equation can be multiplied by 5 and the second equation by 12 to eliminate x.

The THIRD sentence is correct.
The first equation can be multiplied by -12 and the second equation by 5, to eliminate x.
Let's do this:
(1st equation) 5*x + 13*y = 232 / *(-12)
(2nd equation) 12*x + 7*y = 218 / *(5)

When multiplied, we get:
(1st equation) -60*x - 156*y = -2784
(2nd equation) 60*x + 35*y = 1090

Further, first equation now will be obtained when ADDING 1st and 2nd equation together:
(1st equation) -60x+60x-156y+35y=-2784+1090
2nd equation we get from just writing any equation from the beginning, for example, the second one:
(2nd equation) 12*x + 7*y = 218

(1st equation) 0*x - 121*y = -1694
(2nd equation) 12*x + 7*y = 218

(1st equation) -121*y = -1694 [x variable is in this step eliminated]
(2nd equation) 12*x + 7*y = 218

(1st equation) y = 1694/121
(2nd equation) 12*x + 7*y = 218

(1st equation) y = 14
(2nd equation) 12*x + 7*14 = 218

(1st equation) y = 14
(2nd equation) 12*x + 98 = 218

(1st equation) y = 14
(2nd equation) 12*x = 218 - 98

(1st equation) y = 14
(2nd equation) 12*x = 120

(1st equation) y = 14
(2nd equation) x = 120/12

(1st equation) y = 14
(2nd equation) x = 10

So, solution of the system of this two equations is obtained, and it is:
(x,y)=(10,14)