SOLUTION: at noon a plane left austin for lax, 2100 kms away at 500 km per hour. an hour later a jet left lax for austin at 700 kms per hour. at what time did they pass each other?

Algebra ->  Linear-equations -> SOLUTION: at noon a plane left austin for lax, 2100 kms away at 500 km per hour. an hour later a jet left lax for austin at 700 kms per hour. at what time did they pass each other?      Log On


   

Question 800317: at noon a plane left austin for lax, 2100 kms away at 500 km per hour. an hour later a jet left lax for austin at 700 kms per hour. at what time did they pass each other?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+500+%2A1+=+500+ km is the distance the Austin to LAX
plane travels in the hour before the LAX to Austin takes off
Let +d+ = the distance LAX to Austin travels until they meet
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Start a stopwatch when the LAX to Austin plane takes off
Let +t+ = the time in hours for them to pass each other
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Equation for Austin to LAX:
(1) +2100+-+d+-+500+=+500t+
Equation for LAX to Austin:
(2) +d+=+700t++
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Substitute (2) into (1)
(1) +2100+-+700t+-+500+=+500t+
(1) +1200t+=+1600+
(1) +t+=+1.333+
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It was +t+%2B+1+=+2.333+ earlier at noon
when Austin top LAX took off
+.333%2A60+=+20+ minutes
12:00 + 2.333 = 2:20 PM when they passed each other
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check:
(2) +d+=+700t++
(2) +d+=+700%2A1.333+
(2) +d+=+933.333+
and
(1) +2100+-+d+-+500+=+500%2A1.333+
(1) +1600+-+d+=+666.667+
(1) +d+=+933.333+
OK