document.write( "Question 195004: A man made a 70-mile trip at a certain average rate. By increasing his average rate by 5 miles per hour, he traveled 80 miles in the same time that he spent on the 70-mile trip. Find his average rate on each trip. \n" ); document.write( "
Algebra.Com's Answer #146286 by ankor@dixie-net.com(22740)\"\" \"About 
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A man made a 70-mile trip at a certain average rate.
\n" ); document.write( " By increasing his average rate by 5 miles per hour, he traveled 80 miles in
\n" ); document.write( "the same time that he spent on the 70-mile trip. Find his average rate on each trip.
\n" ); document.write( ":
\n" ); document.write( "Let s = the original speed (70 mi trip)
\n" ); document.write( "then
\n" ); document.write( "(s+5) = the speed required for the 80 mi trip
\n" ); document.write( ":
\n" ); document.write( "The two trips have equal time; write a time equation; Time = \"dist%2Fspeed\"
\n" ); document.write( ":
\n" ); document.write( "80 mi time = 70 mi time
\n" ); document.write( "\"80%2F%28%28s%2B5%29%29\" = \"70%2Fs\"
\n" ); document.write( "Cross multiply and you have:
\n" ); document.write( "80s = 70(s+5)
\n" ); document.write( ":
\n" ); document.write( "80s = 70s + 350
\n" ); document.write( ":
\n" ); document.write( "80s - 70s = 350
\n" ); document.write( ":
\n" ); document.write( "10s = 350
\n" ); document.write( "s = \"350%2F10\"
\n" ); document.write( "s = 35 mph original speed
\n" ); document.write( "then
\n" ); document.write( "35 + 5 = 40 mph is the faster speed
\n" ); document.write( ";
\n" ); document.write( ":
\n" ); document.write( "Check solution by finding the time of each trip
\n" ); document.write( "70/35 = 2 hr
\n" ); document.write( "80/40 = 2 hr
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