Question 166131
lets call the distance traveled d(d1) and 1615-d(d2)
fast plane rate of speed lets call f
slow planes rate of speed will call s which equals 1/2f+80--->(2f+160)/2
time = 3 4/5hours or 19/5 hours. Time for plane 1 and plane 2 are equal

so d1=(rate of speed)(time)--->
   d=(f+160/2)(19/5)=(19f+3040)/10

and d2=f(19/5)---->1615-d=19f/5

placing d's value into the d2 equation we have 1615-(19f+3040)/10=19f/5

multiply each term by 10  16150-19f-3040=38f--->57f=13110={{{230}}}

so the speed of the fast plane was {{{230mph}}}
hence the speed of the slow plane was s=1/2(230)+80={{{195mph}}}