SOLUTION: Demarco walks from his house to his friend's house at a constant rate of 5 miles per hour. When he walks back to his house, he walks at a constant rate of 3 miles per hour. Demarco

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Question 1010057: Demarco walks from his house to his friend's house at a constant rate of 5 miles per hour. When he walks back to his house, he walks at a constant rate of 3 miles per hour. Demarco spent 1 hour and 36 minutes in total walking to his friend's house and back home.
What was the distance from Demarco's house to his friends house?
Can you please show how to solve this problem. Thank you so much for your help!
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +d+ = the unknown distance in miles
Let +t+ = time in hrs to get to his friend's house
+1.6+-+t+ = time in hrs to get back from friend's house
( note that I converted 36 min to .6 hrs )
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Equation for walking to friend's house:
(1) +d+=+5t+
Equation for walking back home:
(2) +d+=+3%2A%28+1.6+-+t+%29+
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(2) +d+=+4.8+-+3t+
Substitute (2) into (1)
(1) +d+=+5t+
(1) +4.8+-+3t+=+5t+
(1) +8t+=+4.8+
(1) +t+=+.6+
Plug this back into either equation
(1) +d+=+5%2A.6+
(1) +d+=+3+
The distance is 3 mi
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check:
(2) +d+=+3%2A%28+1.6+-+.6+%29+
(2) +d+=+3%2A1+
(2) +d+=+3+ mi
OK