Part A
Each year, the number of cars in Cory's collection (c) increases by a factor of 1.2. After t years, it will be multiplied by that factor "t" times, so the number of cars will be ...
... c(t) = 15·1.2^t
Each year, the number of cars in Roger's collection (r) increases by 1. After t years, it will have increased by t.
... r(t) = 40 +t
Part B
c(6) = 15·1.2^6 ≈ 44.79 ≈ 45 . . . Cory's cars after 6 years
r(6) = 40 +6 = 46 . . . Roger's cars after 6 years
Part C
The number of cars is very nearly the same after 6 years. A short time into the 7th year, the number of cars will be the same. A graph shows the number of cars will be the same after about 6 years and 2 months.
There are no algebraic methods to solve the equation
... r(t) = c(t)
... 45 +t = 15·1.2^t
It can be solved by trial and error, or graphically, or by other iterative means. The solution is not an integer, but non-integer numbers of cars don't make sense. It isn't exactly clear what a solution would look like.
year 6: Cory ≈ 45 cars; Roger = 46 cars
year 7: Cory ≈ 54 cars; Roger = 47 cars
Between year 6 and year 7 (closer to the beginning of year 7), Cory will have 46 cars, about the same number Roger has at that time.
