The driver run the one way trip a distance of 60*t1 miles ( let t1 is the time for the one way trip)
so we have distance = 60t1
on the wayback driver runs the same distance which is now 50t2 miles. (t2 is the time for return trip)
since the distance is the same we have 60t1 = 50t2 (1) we know that t1 + t2 = 6.6 hours (2)
from (2) we have t2 = 6.6 - t1 replacing this to (1) we get 60t1 = 50(6.6-t1) ->> 60t1=330-50t1 ->>
60t1 + 50t1 = 330 and 110t1 = 330 ->> t1=330/110 = 3 hours substituting where t1 in (2) we get
t2 = 6.6-3 = 3.6 hours