document.write( "Question 142725: A speedboat could travel at 6 times the speed of the current. Thus, it could travel 350 miles downriver in 3 hours more than it took to travel 150 miles upriver. What was the speed of the boast in still water? \n" ); document.write( "
Algebra.Com's Answer #103867 by ptaylor(2198)\"\" \"About 
You can put this solution on YOUR website!
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r\r
\n" ); document.write( "\n" ); document.write( "Let r=rate of the current
\n" ); document.write( "Then 6r=rate of speedboat in still water
\n" ); document.write( "Rate upriver =6r-r=5r
\n" ); document.write( "Rate downriver=6r+r=7r\r
\n" ); document.write( "\n" ); document.write( "Time to travel up river=150/5r
\n" ); document.write( "Time to travel down river =350/7r\r
\n" ); document.write( "\n" ); document.write( "Now we are told the following:
\n" ); document.write( "(150/5r)+3=350/7r simplify
\n" ); document.write( "(30/r)+3=50/r multiply each term by r\r
\n" ); document.write( "\n" ); document.write( "30+3r=50 subtract 30 from each side\r
\n" ); document.write( "\n" ); document.write( "30-30+3r=50-30
\n" ); document.write( "3r=20 divide each side by 3
\n" ); document.write( "r=6 2/3 mph-------------------speed of current
\n" ); document.write( "6r=6*(6 2/3)=40 mph--------------------------speed (rate) of boat in still water\r
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\n" ); document.write( "\n" ); document.write( "CK
\n" ); document.write( "(150/(5*6 2/3))+3=350/(7*6 2/3)) and this equals\r
\n" ); document.write( "\n" ); document.write( "30/(20/3)+3=50/(20/3)
\n" ); document.write( "90/20 +3 =150/20
\n" ); document.write( "4.5+3=7.5
\n" ); document.write( "7.5=7.5\r
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\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor
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