document.write( "Question 89312This question is from textbook
\n" ); document.write( ": If a freight train and an express train leave town 390km apart, traveling towards one another. The freight train travels 30km slower than the express train they pass one another 3hours later. What are their speeds? \n" ); document.write( "
Algebra.Com's Answer #64882 by ptaylor(2198)\"\" \"About 
You can put this solution on YOUR website!
distance(d) equals rate(r) times time(t) or d=rt; t=d/r and r=d/t\r
\n" ); document.write( "\n" ); document.write( "Let r=speed of express
\n" ); document.write( "Then r-30=speed of freight train\r
\n" ); document.write( "\n" ); document.write( "distance express travels=speed of express times 3 hours=3r
\n" ); document.write( "distance freight train travels=speed of freight train times 3 hours=(r-30)*3\r
\n" ); document.write( "\n" ); document.write( "Now we know that as they pass the distance express has traveled plus the distance the freight train has traveled equals 390km. So our equation to solve is:\r
\n" ); document.write( "\n" ); document.write( "3r+3(r-30)=390 get rid of parens\r
\n" ); document.write( "\n" ); document.write( "3r+3r-90=390 add 90 to both sides\r
\n" ); document.write( "\n" ); document.write( "3r+3r-90+90=390+90 collect like terms\r
\n" ); document.write( "\n" ); document.write( "6r=480 divide both sides by 6\r
\n" ); document.write( "\n" ); document.write( "r=80km/hr--------------------------speed of express\r
\n" ); document.write( "\n" ); document.write( "r-30=80-30=50km/hr---------------------speed of freight train\r
\n" ); document.write( "\n" ); document.write( "CK
\n" ); document.write( "3*80+3*50=390
\n" ); document.write( "240+150=390
\n" ); document.write( "390=390\r
\n" ); document.write( "\n" ); document.write( "Hope this helps----ptaylor
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