SOLUTION: Tommy drove his fully-loaded moving van 270 miles to Oklahoma City. With his van empty, Tommy drove 15mi/hr faster and made the return trip in 1.5 hours less. find the speed going.

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Question 40900: Tommy drove his fully-loaded moving van 270 miles to Oklahoma City. With his van empty, Tommy drove 15mi/hr faster and made the return trip in 1.5 hours less. find the speed going.
____mph
Thanks
Found 2 solutions by checkley71, rajagopalan:
Answer by checkley71(8403) (Show Source):
You can put this solution on YOUR website!
270/X=270/(X+15)-1.5 OR 270/X=(270-1.5X-22.5)/(X+15) OR
270(X+15)=X(270-1.5X-22.5) OR 270X+4050=270X-1.5X~2-22.5X OR
4050=-1.5X~2-22.5X OR 1.5X+22.5X-4050=0 OR X+15-2700=0 OR
(X+60)(X-45)=0 OR X=45 MPH GOING.

Answer by rajagopalan(174) (Show Source):
You can put this solution on YOUR website!
Dear Student,
onward journey ..
speed = v
distane =270
time = d/v = 270/v
return journey
speed = v+15
distance = 270
time = 270/(v+15)
difference of two times (270/v) - (270/(v+15)) = 1.5
(270/v) - (270/(v+15)) = (3/2)
(270(v+15)-270v)/v(v+15)=3/2
2(270v+4050-270v)=3(v^2-45v)
8100=3v^2 + 45v
3v^2 + 45v - 8100 =0
divide by 3 both sides
v^2 + 15v + 2700 = 0
v^2 +60v - 45v-2700=0
v(v+60)-45(v+60)=0
(v-45)(v+60)=0
v=45 or -60
excluding negative speed
onward speed v=45
return speed = v+15 = 45+15 = 60
check onward time = 270/45 = 6
return time = 270/60 = 4.5
difference in time = 6-4.5 = 1.5 ..cheers