Question 1085092
Let s and s+10 be the speeds of the truck and the car, respectively. Then:
378/s + 320/s+10=12
378(s+10)+320s=12(sē+10s)
378s+3780+320s=12sē+120s
12sē-578s-3780=0
6sē-289s-1890=0
Using the quadratic formula, we get roots of 54 and -5.833333, so we use 54 mph as the positive speed of the truck.
Also:
(6s+35)(s-54)=0 also gives us roots of 54 and -5.8333333. ☺☺☺☺