The easiest way to solve this is to think of the two towns as being placed on a number line, with Wallace at 0 (zero) and Vernonville at 112. That way, you can write equations that define the position of the cars.
The position of the the first car can then be defined as:
a = (s+10)*t
where a is the position (in miles) of the car, s is the speed of the second car (hence s+10 because it's going 10 mph faster), and t is the time (in hours) since the cars started driving.
The position of the second car could be defined as:
b = 112 - s*t
where b is the position (in miles) of the second car. Notice that we subtract s*t because the car is driving from position 112 towards 0.
We are told that the two cars meet after 1 hour and 36 minutes. In decimal form, this would be 1.6 hours. When the cars meet, they have the same position, hence:
a=b
which gives:
(s+10)*t = 112 - s*t
subbing in 1.6 for t gives:
(s+10)*1.6 = 112 - s*1.6
and finally, solving for s gives:
1.6s + 160 = 112 - 1.6s
3.2s +16 = 112
3.2s = 96
s = 30
Hence, the slower car (the one starting in Vernonville) travels at 30mph, and the other travels 10mph faster, going 40mph.
