document.write( "Question 1048246: paul traveled to lexington at 70 mph. on the way back he averages 55 mph. the way back was 86 miles longer and took 2.6 hours longer to drive. how long was the scenic route from Lexington? \n" ); document.write( "
Algebra.Com's Answer #663841 by Alan3354(69443)![]() ![]() You can put this solution on YOUR website! paul traveled to lexington at 70 mph. on the way back he averages 55 mph. the way back was 86 miles longer and took 2.6 hours longer to drive. how long was the scenic route from Lexington? \n" ); document.write( "---------------- \n" ); document.write( "d = r*t \n" ); document.write( "r = rate going = 70 mi/hr, d = distance going, t = time going \n" ); document.write( "--- \n" ); document.write( "d = 70t \n" ); document.write( "---- \n" ); document.write( "d + 86 = 55*(t + 2.6) \n" ); document.write( "--- \n" ); document.write( "70t + 85 = 55t + 143 \n" ); document.write( "15t = 58 \n" ); document.write( "t = 58/15 \n" ); document.write( "================ \n" ); document.write( "scenic d = 55*(58/15 + 2.6) \n" ); document.write( "= 11*58/3 + 143 \n" ); document.write( "=~ 355.67 miles \n" ); document.write( " |