document.write( "Question 726238: Bill sets out to drive from Swansea to London. An hour after he'd started, he encountered heavy traffic and he was only able to drive at three-fifths of his former speed. He eventually arrived in London two hours later than he'd expected, and worked out that if he'd not met traffic until fifty miles further on he would have arrived in London forty minutes sooner. How far is it from Swansea to London? \n" ); document.write( "
Algebra.Com's Answer #444553 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! Bill sets out to drive from Swansea to London. \n" ); document.write( " An hour after he'd started, he encountered heavy traffic and he was only able to drive at three-fifths of his former speed. \n" ); document.write( "let s = his normal speed \n" ); document.write( "then \n" ); document.write( ".6s = his heavy traffic speed \n" ); document.write( "and \n" ); document.write( "let d = the distance from Swansea to London \n" ); document.write( " He eventually arrived in London two hours later than he'd expected, \n" ); document.write( "Write a time equation of this scenario \n" ); document.write( "1 + \n" ); document.write( "multiply by .6s, resulting in: \n" ); document.write( ".6s + (d-s) = .6d + .6s(2) \n" ); document.write( ".6s - s + d = .6d + 1.2s \n" ); document.write( "-.4s + d = .6d + 1.2s \n" ); document.write( "d - .6d = 1.2s + .4s \n" ); document.write( ".4d = 1.6s \n" ); document.write( "d = \n" ); document.write( "d = 4s \n" ); document.write( ": \n" ); document.write( "and worked out that if he'd not met traffic until fifty miles further on he would have arrived in London forty minutes sooner. \n" ); document.write( "That means he would only be 1 hr and 20 min late, 4/3 hr \n" ); document.write( "1 + \n" ); document.write( "Multiply by 3s \n" ); document.write( "3s + 3(50) + 5(d-s-50) = 3d + 4s \n" ); document.write( "3s + 150 + 5d - 5s - 250 = 3d + 4s \n" ); document.write( "combine like terms \n" ); document.write( "3s - 5s - 4s + 150 - 250 = 3d - 5d \n" ); document.write( "-6s - 100 = -2d \n" ); document.write( "6s + 100 = 2d, mult by -1 \n" ); document.write( "simplify divide by 2 \n" ); document.write( "3s + 50 = d \n" ); document.write( "replace d with 4s \n" ); document.write( "3s + 50 = 4s \n" ); document.write( "50 = 4s - 3s \n" ); document.write( "s = 50 mph is the speed \n" ); document.write( "then \n" ); document.write( "4(50) = 200 mi from Swansea to London\r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |