Question 220073

Distance(d) equals Rate(r) times Time(t) or d=rt;  r=d/t and t=d/r

Let r=Alberto's average rate of speed on the outbound trip
Then (r-9.8)=Alberto's average rate of speed on the inbound trip
Distance travelled on the outbound trip=(r)*3
Distance travelled on the inbound trip=(r-9.8)*4
Now we know that the above two distances are equal, so our equation to solve is:
3r=4(r-9.8)  get rid of parens
3r=4r-39.2  subtract 4r from each side
3r-4r=4r-4r-39.2 collect like terms
-r=-39.2 mph or
r=39.2 mph---Alberto's rate of speed on the outbound trip
r-9.4=39.2-9.8= 29.4 mph---Alberto's average speed on the inbound trip

CK
Distance travelled on outbound trip=39.2*3=117.6 mi
Distance travelled on the inbound trip=29.4*4=117.6 mi

Hope this helps---ptaylor