document.write( "Question 358424: Andre traveled from his home to the office at an average rate of 25 kph. By traveling 5 kph faster, he took 30 min less to return home. How far is his office from his home? \n" ); document.write( "
Algebra.Com's Answer #255826 by checkley77(12844)![]() ![]() ![]() You can put this solution on YOUR website! D=25T \n" ); document.write( "D=30(T-.5)) \n" ); document.write( "25T=30(T-.5) \n" ); document.write( "25T=30T-15 \n" ); document.write( "25T-30T=-15 \n" ); document.write( "-5T=-15 \n" ); document.write( "T=-15/-5 \n" ); document.write( "T=3 HOURS FOR THE 25MPH. TRIP. \n" ); document.write( "25*3=75 MILES FROM HOME TO OFFICE \n" ); document.write( "30(3-.5)=30*2.5=75 DITTO. \n" ); document.write( " \n" ); document.write( " |