document.write( "Question 121717: A boy has planned a three-hour bicycle ride. After biking at a rate of 14 mph for awhile, the bike breaks down and he rides back with his father at a rate of thirty-five mph. How far did the boy ride his bicycle if he returns one hour after he left? \n" ); document.write( "
Algebra.Com's Answer #89358 by ankor@dixie-net.com(22740)\"\" \"About 
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A boy has planned a three-hour bicycle ride. After biking at a rate of 14 mph for awhile, the bike breaks down and he rides back with his father at a rate of thirty-five mph. How far did the boy ride his bicycle if he returns one hour after he left?
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\n" ); document.write( "Let d = distance ridden on the bike
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\n" ); document.write( "Write two time equations with the information we have: Time = Distance/speed
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\n" ); document.write( "time on the bike = \"d%2F14\"
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\n" ); document.write( "Time in the car = \"d%2F35\"; (distance on the bike = return distance in the car)
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\n" ); document.write( "Time on bike + time in car = 1 hr
\n" ); document.write( "\"d%2F14\" + \"d%2F35\" = 1
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\n" ); document.write( "Multiply equation by 14*35 = 490, to get rid of the denominators
\n" ); document.write( "490*\"d%2F14\" + 490*\"d%2F35\" = 490(1)
\n" ); document.write( "Cancel the denominators and you have:
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\n" ); document.write( "35d + 14d = 490
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\n" ); document.write( "49d = 490
\n" ); document.write( "d = \"490%2F10\"
\n" ); document.write( "d = 10 miles
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\n" ); document.write( "Check solution by finding the total time
\n" ); document.write( "10/14 + 10/35 =
\n" ); document.write( ".714 + .286 = 1 hr, confirms our solution
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\n" ); document.write( "Did this make sense to you? Any questions?
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