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)\"\" \"About 
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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.
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