document.write( "Question 120468: a passenger train can travel 325 mi in the same time a freight train takes to travel 200 mi. if the speed of the passenger train is 25 mi/h faster than the speed of the freiht train, find the speed of each. \n" ); document.write( "
Algebra.Com's Answer #88300 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
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\n" ); document.write( "\n" ); document.write( "Let r=rate (speed) of the freight train
\n" ); document.write( "Then r+25=rate of the passenger train
\n" ); document.write( "Time for freight train to travel 200 mi=200/r
\n" ); document.write( "Time for passenger train to travel 325 mi=325/(r+25)\r
\n" ); document.write( "\n" ); document.write( "Now we are told that the above two times are equal, so:\r
\n" ); document.write( "\n" ); document.write( "200/r=325/(r+25) multiply each side by r(r+25){or cross-multiply}\r
\n" ); document.write( "\n" ); document.write( "200(r+25)=325r get rid of parens\r
\n" ); document.write( "\n" ); document.write( "200r+5000=325r subtract 200r from both sides\r
\n" ); document.write( "\n" ); document.write( "200r-200r+5000=325r-200r collect like terms\r
\n" ); document.write( "\n" ); document.write( "125r=5000 divide both sides by 125
\n" ); document.write( "r=40 mph-----------------------------------speed of freight train\r
\n" ); document.write( "\n" ); document.write( "r+25=40+25=65 mph-----------------------------speed of passenger train\r
\n" ); document.write( "\n" ); document.write( "CK\r
\n" ); document.write( "\n" ); document.write( "200/40=325/65
\n" ); document.write( "5=5\r
\n" ); document.write( "\n" ); document.write( "Hope this helps---ptaylor
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