document.write( "Question 101286: On Friday morning Ricardo drove from his house to the beach in 4 hours. In coming back on Sunday afternoon, heavy traffic slowed his speed by 7 mi/h and the trip took 5 hours. What was his average speed (rate) in each direction? \n" ); document.write( "
Algebra.Com's Answer #73854 by ankor@dixie-net.com(22740)\"\" \"About 
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from his house to the beach in 4 hours. Back, traffic slowed his speed by 7 mi/h and the trip took 5 hours. What was his average speed (rate) in each direction?\r
\n" ); document.write( "\n" ); document.write( "Let t = speed to the beach
\n" ); document.write( "then
\n" ); document.write( "(t-7) = speed back from the beach
\n" ); document.write( ":
\n" ); document.write( "Distance there and back is the same so write a distance equation:
\n" ); document.write( "Distance = time * speed
\n" ); document.write( ":
\n" ); document.write( "from the beach dist = to the beach dist
\n" ); document.write( "5(t-7) = 4t
\n" ); document.write( ":
\n" ); document.write( "5t - 35 = 4t
\n" ); document.write( ":
\n" ); document.write( "5t - 4t = + 35
\n" ); document.write( ":
\n" ); document.write( "t = 35 mph to, and 28 mph back
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "Check solution by confirming the distances are the same:
\n" ); document.write( "4 * 35 = 140
\n" ); document.write( ":
\n" ); document.write( "5 * 28 = 140
\n" ); document.write( ":
\n" ); document.write( "That's pretty easy, right?
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