SOLUTION: This is from a text book and I cannot figure how to set it up and answer it. CAR TRIP. During the first part of a trip, Trudy's Honda traveled 120 mi at a certain speed. Trudy t

Question 187209This question is from textbook
: This is from a text book and I cannot figure how to set it up and answer it.
CAR TRIP. During the first part of a trip, Trudy's Honda traveled 120 mi at a certain speed. Trudy then drove another 100 miles at a speed that was 10 mph slower. If the total time of Trudy's trip was 4 hr, what was her speed on each part of the trip? I am stumped at the moment. Thank you for your help. This question is from textbook

Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!

WOULD IT HELP IF I SET IT UP AND YOU ANSWER IT? OK. I'LL SET IT UP.
Distanced(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
Let r=her speed during the first part of the trip
Then r-10= her speed during the second part of the trip
time for first part of trip=120/r
time for second part of trip=100/(r-10)
And we are told that the sum of the above two times equals 4 hrs, so:
120/r +100/(r-10)=4 multiply each term by r(r-10)
120(r-10)+100r=4r(r-10) get rid of parens
120r-1200+100r=4r^2-40r subtract 4r^2 from and add 40r to both sides
120r-1200+100r-4r^2+40r=0 collect like terms
-4r^2+260r-1200=0 divide each term by -4
r^2-65r+300=0 quadratic in standard form and it can be factored:
(r-60)(r-5)=0
By inspection, we can see that 5 is no good because when we subtract 10 from it, we get a negative speed and that's not good.
I think you can take it from here
Hope this helps---ptaylor

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