document.write( "Question 1088662: An excursion boat travels up river to a landing than returns to its starting point. The trip upriver takes 2.3 hours, and the trip back takes 1.8 hours. If the average speed on the return trip is 5 mph faster than the trip up river, what is the boat’s speed up river? Hint: Distance = rate * time. Draw a table.ate * time. Draw a table. \n" ); document.write( "
Algebra.Com's Answer #703008 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
Let \"+s+\" = the speed going upriver in mi/hr
\n" ); document.write( "\"+s+%2B+5+\" = the speed going downriver in mi/hr
\n" ); document.write( "------------------
\n" ); document.write( "(1) \"+d+=+s%2A2.3+\"
\n" ); document.write( "and
\n" ); document.write( "(2) \"+d+=+%28+s+%2B+5+%29%2A1.8+\"
\n" ); document.write( "--------------------------
\n" ); document.write( "(2) \"+d+=+1.8s+%2B+9+\"
\n" ); document.write( "Plug (1) into (2)
\n" ); document.write( "(2) \"+2.3s+=+1.8s+%2B+9+\"
\n" ); document.write( "(2) \"+.5s+=+9+\"
\n" ); document.write( "(2) \"+s+=+18+\"
\n" ); document.write( "------------------
\n" ); document.write( "The speed upriver is 18 mi/hr
\n" ); document.write( "-----------------
\n" ); document.write( "To make a table, you have:
\n" ); document.write( "[ upriver speed, s ], [ downriver speed, s+5 ],
\n" ); document.write( "[ upriver time ], [ downriver time ]
\n" ); document.write( "[ upriver distance, d ] = [ downriver distance, d ]\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "
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