SOLUTION: a. Mike and Scott live 110 miles from each other. They frequently meet for lunch at a restaurant that is between Mike’s house and Scott’s house. Leaving at the same time from the

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: a. Mike and Scott live 110 miles from each other. They frequently meet for lunch at a restaurant that is between Mike’s house and Scott’s house. Leaving at the same time from the      Log On

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Question 607157: a. Mike and Scott live 110 miles from each other. They frequently meet for lunch at a restaurant that is between Mike’s house and Scott’s house. Leaving at the same time from their respective houses, Mike takes 1 hour and 30 minutes and Scott takes 1 hour and 15 minutes to get to the restaurant. If they each drive at the same speed,
i. Determine their speed
ii. How far from Scott’s house is the restaurant?
Thank you!

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +d+ = distance from Scott's house to restaurant
+110+-+d+ = distance from Mike's house to restaurant
+s+ = speed for both Mike and Scott
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Scott's equation:
(1) +d+=+s%2A1.25+
Mike's equation:
(2) +110+-+d+=+s%2A1.5+
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Substitute (1) into (2)
(2) +110+-+1.25s+=+1.5s+
(2) +2.75s+=+110+
(2) +s+=+40+ mi/hr
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Both speeds are 40 mi/hr
(1) +d+=+s%2A1.25+
(1) +d+=+40%2A1.25+
(1) +d+=+50+
Scott's house is 50 mi from restaurant
check:
(2) +110+-+d+=+s%2A1.5+
(2) +110+-+d+=+40%2A1.5+
(2) +110+-+d+=+60+
(2) +d+=+110+-+60+
(2) +d+=+50+
OK