Question 401141
Let r = speed at the 1st part of trip ==> time for 1st part = 48/r
==> r - 5 = speed at 2nd part of trip ==> time for 2nd part = 19/(r-5).
both from the formula t = d/r = distance/rate.

Then {{{48/r + 19/(r-5) = 3}}}
<==> 48(r-5) + 19r = 3r(r-5)
<==> {{{48r - 240 + 19r = 3r^2 - 15r}}}
<==> {{{3r^2 - 82r + 240 = 0}}} <==> (r - 24)(3r - 10) = 0
==> r = 24, 10/3
Eliminate r = 10/3, because it would make r - 5 negative.
Hence the speed in the 1st part is 24 mph, and the speed in the 2nd part is 19 mph.