document.write( "Question 475494: Cheryl left home on a business trip in her car and averaged 50mph. The next day she started to return at an average rate of 38 mph. After driving 2 hours longer than she spent on her outward trip. She discovered that she was still 14 miles from home. How far from home had she traveled? \n" ); document.write( "
Algebra.Com's Answer #326062 by ankor@dixie-net.com(22740)\"\" \"About 
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Cheryl left home on a business trip in her car and averaged 50mph.
\n" ); document.write( " The next day she started to return at an average rate of 38 mph.
\n" ); document.write( " After driving 2 hours longer than she spent on her outward trip, she discovered that she was still 14 miles from home.
\n" ); document.write( " How far from home had she traveled?
\n" ); document.write( ":
\n" ); document.write( "Let t = travel time on the 1st day
\n" ); document.write( "then
\n" ); document.write( "(t+2) = travel time on the 2nd day to get 14 mi from home
\n" ); document.write( ":
\n" ); document.write( "Write a distance equation, dist = speed * time
\n" ); document.write( ":
\n" ); document.write( "1st day dist = 2nd day dist + 14 mi
\n" ); document.write( "50t = 38(t+2) + 14
\n" ); document.write( "50t = 38t + 76 + 14
\n" ); document.write( "50t - 38t = 90
\n" ); document.write( "12t = 90
\n" ); document.write( "t = \"90%2F12\"
\n" ); document.write( "t = 7.5 hrs
\n" ); document.write( ":
\n" ); document.write( "Find the distance:
\n" ); document.write( "50 * 7.5 = 375 mi
\n" ); document.write( ":
\n" ); document.write( "Check the dist
\n" ); document.write( "38(7.5+2) + 14
\n" ); document.write( "361 + 14 = 375 \n" ); document.write( "
\n" );