To solve this you need to use simultaneous equations (or that's the name that I was taught). We write one for each situation. I'm using v as the variable for van and "b" for bus.
11v + 12b = 412
9v + 2b = 126 (I'll solve this one for b)
2b = 126 - 9v
b = (126 - 9v)/2
Now I'll substitute that value for b into the other equation. This leaves me with only the v variable. (algebra commentary for each step)
11v + 12((126 - 9v)/2) = 412 (substitute)
11v + (1512-108v)/2 = 412 (multiply the 12 through)
11v +756 - 54v = 412 (divide by 2)
-43v + 756 = 412 (combine "v" terms)
-43v = -344 (subtract 756 from both sides)
v = 8 (divide both sides by -43)
Now we know how many people a van holds. We can put that value in place of "v" in either expression in order to solve for b.
9(8) +2b = 126 (put 8 in for v)
72 + 2b = 126 (multiply)
2b = 54 (subtract 72 from both sides)
b = 27 (divide both sides by 2)
It's a strange bus that holds 27 students, but hey, it's a math problem. We can check that these values work by plugging them both into the other equation.
11(8) + 12(27) = 412 It works