#### 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.