document.write( "Question 440961: Wilbur drove to the town hall and back. The trip there took 1.8 hours and the trip back took 2.2 hours. He averaged 9 mph faster on the trip there than on the return trip. What was Wilbur's average speed on the outbound trip? \n" ); document.write( "
Algebra.Com's Answer #304486 by jorel1380(3719) You can put this solution on YOUR website! 1.8(x+9)=2.2x \n" ); document.write( "1.8x+16.9=2.2x \n" ); document.write( "16.9=.4x \n" ); document.write( "16.9/.4=x \n" ); document.write( "x=40.5 \n" ); document.write( "x+9=49.5 \n" ); document.write( "Wilbur's average speed outgoing was 49.5mph.. \n" ); document.write( " |