The key here is the equation d = rt, which describes the relationship between distance, rate, and time.
The equation can be applied to the two different situations. It can be applied to the 65 mi bicycle trip and the 185 mi car trip. What do these two situations have in common?
They both take the same amount of time.
To start, let's turn the words into algebra. We need to define our variables.
Let x = the distance driven, y = the distanced biked,
b = the rate of biking, c = the rate of car-driving
d = rt can be solved for t as t = d/r
Because the time in both situations is the same, we can write
x/c = y/b
So far so good, except now we have an equation with 2 unknowns. But we know that c = 40 + b (see problem description), so we can put that in instead. We have 1 equation now to solve with 1 unknown.
x/(40 + b) = y/b
Multiply both sides by the product of the denominators (cross multiply):
bx = y(40 + b)
Move the "b's" to one side of the equation.
bx = 40y + by
bx - by = 40y
Turn the two "b's" into one "b" by factoring.
b(x - y) = 40y
b = 40y/(x - y)
Plug in actual numbers.
b = 40*65/(185 - 65) = 21.67 mi/h
c = 40 + b = 40 + 21.67 = 61.67 mi/h
Check.
Calculate t for both situations.
t = y/b = 65/21.67 = 3 h
t = x/c = 185/61.67 = 3 h
