SOLUTION: At noon a private plane left Austin for Los Angeles,2100km away, flying at 500km/h. One hour later a jet left Los Angeles for Austin at 700km/h. At what time did they pass each oth
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Question 113760This question is from textbook Algebra Structure and method
: At noon a private plane left Austin for Los Angeles,2100km away, flying at 500km/h. One hour later a jet left Los Angeles for Austin at 700km/h. At what time did they pass each other?
Here is what I have so far.I am not sure how to do this
2100=500(t+60)-700(t-60)
This question is from textbook Algebra Structure and method
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
At noon a private plane left Austin for Los Angeles,2100km away, flying at 500km/h. One hour later a jet left Los Angeles for Austin at 700km/h. At what time did they pass each other?
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Private plane DATA:
Rate = 500 km/h ; time = x hrs; distance = 500x km
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Jet DATA:
Rate = 700 km/h ; time = x-1 hrs ; distance = 700(x-1)
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EQUATION:
private distance + jet distance = 2100 km
500x + 700(x-1) = 2100
5x + 7x-7 = 21
12x = 28
x= 7/3 hrs = 2 hrs 20 minutes
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They passed at noon + 2 hrs 20 min or 2:20 PM
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Cheers,
Stan H.
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