SOLUTION: Jen recently rode her bike to visit her friend who lives 12 miles away. On her way there, her average speed was 5 miles per hour faster than on her way home. If Jen spent a total o

Algebra ->  Rate-of-work-word-problems -> SOLUTION: Jen recently rode her bike to visit her friend who lives 12 miles away. On her way there, her average speed was 5 miles per hour faster than on her way home. If Jen spent a total o      Log On


   

Question 1057791: Jen recently rode her bike to visit her friend who lives 12 miles away. On her way there, her average speed was 5 miles per hour faster than on her way home. If Jen spent a total of 2 hours biking, find the two rates.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +s+ = her average speed going back home
+s+%2B+5+ = her average speed going to friend's house
Let +t+ = her time in hrs going back home
+2+-+t+ = her time in hrs going to friend's house
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Equation for going to friend's house:
(1) +12+=+%28+s+%2B+5+%29%2A%28+2+-+t+%29+
Equation for going back home:
(2) +12+=+s%2At+
--------------------------------
(2) +t+=+12%2Fs+
Plug this result back into (1)
(1) +12+=+%28+s+%2B+5+%29%2A%28+2+-+12%2Fs+%29+
(1) +12+=+2s+%2B+10+-+12+-+60%2Fs+
(1) +14+=+2s+-+60%2Fs+
(1) +14s+=+2s%5E2+-+60+
(1) +2s%5E2+-+14s+-+60+=+0+
(1) +s%5E2+-+7s+-+30+=+0+
(1) +%28+s+-+10+%29%2A%28+s+%2B+3+%29+=+0+ ( by looking at it )
(1) +s+=+10+ ( can't use the negative solution )
and
+s+%2B+5+=+15+
Going to friend's house: 15 mi/hr
Going back home: 10 mi/hr
-----------------------------------
check:
(2) +12+=+s%2At+
(2) +12+=+10t+
(2) +t+=+1.2+ hrs
and
(1) +12+=+%28+s+%2B+5+%29%2A%28+2+-+t+%29+
(1) +12+=+15%2A%28+2+-+t+%29+
(1) +12+=+30+-+15t+
(1) +15t+=+30+-+12+
(1) +15t+=+18+
(1) +t+=+1.2+ hrs
OK