let

d = 41 miles the trip traveled in one direction

v1 = the speed at the first part of the trip, v1>0

t1 = the time spent of the first part of the trip

v2 = the speed of the return trip, 0<v2<v1

t2 = the time spent of the return trip

t1 + t2 = 14 hr the total time of the trip

t2 = 14 - t1

the return trip is made at a speed that is 6 mph slower

v2 = v1 - 4

since speed = distance/time we have distance = speed*time

v1*t1 = d

v1*t1 = 41

v2*t2 = d

v2*t2 = 41

if we plug t2 = 14 - t1 and v2 = v1 - 4 in the last equation we get

(v1 - 4)(14 - t1) = 41

by solving the system of equations

v1*t1 = 41 => t1 = 41/v1 plug in

(v1 - 4)(14 - t1) = 41

we get

(v1 - 4)(14 - 41/v1) = 41

by solving the equation we find 2 solutions

v1 = 8.47 mph => v2 = v1 - 4 = 8.47 - 4 = 4.47 mph

v1` = 1.38 => v2 = 1.38 - 4 < 0 thus v1` is not a solution

the bicyclist's average speed of the first part is 8.47 mph and the average speed of the return trip is 4.47 mph.