document.write( "Question 6642: mr. west drove his car from home to chicago at the rate of 40 mph. amd returned at a rate of 45mph. if his time going exceed his time returning by 30 minutes, find his time going and his time returning \n" ); document.write( "
Algebra.Com's Answer #3620 by prince_abubu(198)![]() ![]() ![]() You can put this solution on YOUR website! We're going to use rate * time = distance in this situation. Before we go on, we must keep in mind that the distance for both to and from trips is the same.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Let's nail down the trip home first.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "Distance from Chicago to home would then be \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Let's nail down the trip to Chicago then.\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "Distance from home to Chicago would have to be \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Since the rates and times HAVE TO equal the same distance, we can set them equal to each other.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " |