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 #304489 by josmiceli(19441) $\"\"$ $\"About$

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Let $\"s\"$ = his average speed for the return trip.
\n" ); document.write( "given:
\n" ); document.write( "To town hall:
\n" ); document.write( "(1) $\"+d+=+%28s+%2B+9%29%2A1.8+\"$
\n" ); document.write( "The trip back:
\n" ); document.write( "(2) $\"+d+=++s%2A2.2+\"$
\n" ); document.write( "Set (1) equal to (2)
\n" ); document.write( " $\"+%28s+%2B+9%29%2A1.8+=+2.2s+\"$
\n" ); document.write( " $\"+1.8s+%2B+16.2+=+2.2s+\"$
\n" ); document.write( " $\"+.4s+=+16.2+\"$
\n" ); document.write( "Trip back:
\n" ); document.write( " $\"+s+=+40.5+\"$
\n" ); document.write( "To town hall:
\n" ); document.write( " $\"+s+%2B+9+=+49.5+\"$ \n" ); document.write( "

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