Answer
He is going 16 mph on the way out and 12 mph on the way back.
Work Step by Step
Recall that velocity is equal to change in distance over change in time. We call t the time it takes the on the way back. Thus, we obtain:
$\frac{40}{t-1} = \frac{42}{t} +4$
We create common denominators and solve for t:
$ \frac{40t}{t(t-1)} = \frac{42(t-1)}{t(t-1)} + \frac{4t(t-1)}{t(t-1)}$
Since the denominators are the same, we cancel them out and solve. (Recall, t cannot be negative.)
$40t = 42t -42 +4t^2 -4t \\ 0 = 4t^2 -2t -42 \\ (2t-7)(t+3) \\ t=3.5$
Thus, on the way back:
$ v = 42/3.5 = 12$
This means that he was going 16 mph on the way out.
