document.write( "Question 206828: a couple attends an out of town wedding. the trip to the wedding takes 3 hrs the wedding and reception takes 2 hrs and as a result of the celebration the trip home takes 4 hrs at a rate of 15 mph less than the going rate. what was the couples tate of travel on the way home.\r
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document.write( "just overwhelmed dont even know how to approach the problem dont even know how to declare and variable, thank you \n" );
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Algebra.Com's Answer #156513 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! the trip to the wedding takes 3 hrs the wedding and reception takes 2 hrs and as \n" ); document.write( " a result of the celebration the trip home takes 4 hrs at a rate of 15 mph less \n" ); document.write( " than the going rate. what was the couples rate of travel on the way home. \n" ); document.write( ": \n" ); document.write( "let r = rate of travel home \n" ); document.write( "then \n" ); document.write( "(r+15) = rate of travel to the wedding (15 mph faster) \n" ); document.write( ": \n" ); document.write( "The time of the wedding and reception are not relevant to the problem \n" ); document.write( ": \n" ); document.write( "The trip there and trip back are equal distance \n" ); document.write( ": \n" ); document.write( "Write a distance equation: Dist = time * rate \n" ); document.write( ": \n" ); document.write( "4r = 3(r+15) \n" ); document.write( ": \n" ); document.write( "4r = 3r + 45 \n" ); document.write( ": \n" ); document.write( "4r - 3r = 45 \n" ); document.write( ": \n" ); document.write( "r = 45 mph hr on the return trip \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Check solution by finding the distances (should be equal) \n" ); document.write( "outbound trip: 45 + 15 = 60 mph \n" ); document.write( "3(60) = 180 mi \n" ); document.write( "4(45) = 180 mi \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "Not that hard, right? \n" ); document.write( " \n" ); document.write( " |