document.write( "Question 334331: John drove to a distant city in 5 hours. When he returned, there was less traffic and the trip took only 3 hours. If John averaged 26 mph faster on the return trip, how fast did he drive each way?\r
\n" ); document.write( "\n" ); document.write( "It seems that something is missing from the problem. Help please. \n" ); document.write( "
Algebra.Com's Answer #239750 by ankor@dixie-net.com(22740)\"\" \"About 
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John drove to a distant city in 5 hours.
\n" ); document.write( " When he returned, there was less traffic and the trip took only 3 hours.
\n" ); document.write( " If John averaged 26 mph faster on the return trip, how fast did he drive each way?
\n" ); document.write( ":
\n" ); document.write( "We don't know the distance of the trips, but we know they are equal.
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\n" ); document.write( "Let s = speed to the city
\n" ); document.write( "then
\n" ); document.write( "(s+26 = return speed
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\n" ); document.write( "Write a distance equation: dist = time * speed
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\n" ); document.write( "To dist = return dist
\n" ); document.write( "5s = 3(s+26)
\n" ); document.write( "5s = 3s + 78
\n" ); document.write( "5s - 3s = 78
\n" ); document.write( "2s = 78
\n" ); document.write( "s = \"78%2F2\"
\n" ); document.write( "s = 39 mph to the city
\n" ); document.write( "and
\n" ); document.write( "39 + 26 = 65 mph back
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\n" ); document.write( "We can confirm our solution by finding the distances now
\n" ); document.write( "5*39 = 195 mi
\n" ); document.write( "3*65 = 195 mi also
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\n" ); document.write( "Pretty simple, right?\r
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