mr. Duncan traveled to a city 180 miles from him home to attend a meeting. Due to car trouble, his average speed returning was 13 miles less than his speed going. If the total time for the road trip was 7 hours, at what rate of speed did he travel to the city? (round to nearest tenth)
It was 7 hours total round trip?
yes
here is a guess:
\[\frac{miles}{hour} * hour = miles\] so\[hour = \frac{miles}{\frac{miles}{hour}}\]
so the total time is that for each leg of the journey:
\[7 = \frac{180}{x} + \frac{180}{x - 13}\]
solve for x. not sure if this makes sense, even to me :)
a better way to restate what i first said: \[average speed * time = distance\]
okk. So i got 30420 when I did that. ha. I'm sure what you did makes sense. I just am lost with this problem a bit. Did you get a solution at all?
hmm you may have made a mistake with your numbers. let me try and step through it. i am checking my answer here: http://www.wolframalpha.com/input/?i=7+%3D+180%2Fx+%2B+180%2F(x-13) it gives approx 59mph. also gives approx 6mph as well, so i may be wrong.
ohh hmm. I wonder which one it is. According to where you checked it the answer is 59 mph?
there are actually two solutions listed there, which seems indicative of something wrong. but I'm not too sure about that. I would guess 59 because if it was 6mph, then on the way home he would have been going 6-13 mph which is impossible.
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