SOLUTION: Two planes travel toward each other from cities that are 1615 kilometers apart. The two planes passed each other in the air after 3 hours and 48 minutes. The slower plane was trave

Question 166131: Two planes travel toward each other from cities that are 1615 kilometers apart. The two planes passed each other in the air after 3 hours and 48 minutes. The slower plane was traveling at a rate of speed that was 80 k/h more than half the rate of speed of the faster plane. Assuming they left at the same time, at what rate of speed was each plane traveling?
Answer by Mathtut(3670) (Show Source): You can put this solution on YOUR website!
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=
so the speed of the fast plane was
hence the speed of the slow plane was s=1/2(230)+80=
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