SOLUTION: a man rode a bike fot 12 miles and then hiked an addiyional 8 miles. the time of the trip was 5 hrs. if his rate when rifing the bike was 10 miles per hr fastr than his walking rat

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Question 751549: a man rode a bike fot 12 miles and then hiked an addiyional 8 miles. the time of the trip was 5 hrs. if his rate when rifing the bike was 10 miles per hr fastr than his walking rate what was each rate?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +s+ = his speed in mi/hr walking
+s+%2B+10+ = his speed in mi/hr riding the bike
Let +t+ = his time in hrs spent walking
+5+-+t+ = his time in hrs spent riding bike
-----------------
Equation when riding bike:
(1) +12+=+%28+s+%2B+10+%29%2A%28+5+-+t+%29+
Equation when walking:
(2) +8+=+s%2At+
--------------
(1) +12+=+5s+%2B+50+-+s%2At+-+10t+
(1) +5s+-+s%2At+=+10t+%2B+12+-+50+
(1) +s%2A%28+5+-+t+%29+=+10t+-+38+
and
(2) +8+=+s%2At+
(2) +t+=+8%2Fs+
Substitute this into (1)
(1) +s%2A%28+5+-+8%2Fs+%29+=+10%2A%288%2Fs%29+-+38+
(1) +5s+-+8+=+80%2Fs+-+38+
Multiply both sides by +s+
(1) +5s%5E2+-+8s+=+80+-+38s+
(1) +5s%5E2+%2B+30s+=+80+
(1) +s%5E2+%2B+6s+=+16+
complete the square
(1) +s%5E2+%2B+6s+%2B+%286%2F2%29%5E2+=+16+%2B+%286%2F2%29%5E2+
(1) +s%5E2+%2B+6s+%2B+9+=+16+%2B+9+
(1) +%28+s+%2B+3+%29%5E2+=+5%5E2+
(1) +s+%2B+3+=+5+
(1) +s+=+2+
and
+s+%2B+10+=+12+
His speed in mi/hr walking is 2 mi/hr
His speed in mi/hr riding the bike is 12 mi/hr
check:
(2) +8+=+s%2At+
(2) +t+=+8%2F2+
(2) +t+=+4+ hrs
and
(1) +12+=+%28+s+%2B+10+%29%2A%28+5+-+t+%29+
(1) +12+=+%28+2+%2B+10+%29%2A%28+5+-+t+%29+
(1) +12+=+12%2A%28+5+-+t+%29+
(1) +12%2F12+=+5+-+t+
(1) +1+-+5+=+-t+
(1) +-t+=+-4+
(1) +t+=+4+ hrs
OK