document.write( "Question 1099384: A train, traveling at 50 mph., starts 1 and 1/3 hrs. after a slower train, and in 2 hrs., overtakes it. Calculate rate of the slower train.\r
\n" ); document.write( "\n" ); document.write( "Let x = rate of slower train.\r
\n" ); document.write( "\n" ); document.write( "rate * time = distance
\n" ); document.write( "rate = distance / rate\r
\n" ); document.write( "\n" ); document.write( "Faster train: 50 * (t - 1.33)= 50t - 66.5\r
\n" ); document.write( "\n" ); document.write( "Slower: x * t = xt.\r
\n" ); document.write( "\n" ); document.write( "Not sure how to continue. Non-homework.
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Algebra.Com's Answer #713825 by josgarithmetic(39618)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "               SPEED       TIME        DISTANCE\r\n" );
document.write( "FAST           50          2             d\r\n" );
document.write( "SLOW           r           2+1&1/3       d\r\n" );
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\n" ); document.write( "\n" ); document.write( "FAST train data will give the distance.
\n" ); document.write( "SLOW train data will allow to find speed r of the slow train.\r
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\n" ); document.write( "\n" ); document.write( "\"r%282%2B1%261%2F3%29=50%2A2\"\r
\n" ); document.write( "\n" ); document.write( "-\r
\n" ); document.write( "\n" ); document.write( "\"r%283%261%2F3%29=100\"\r
\n" ); document.write( "\n" ); document.write( "\"%2810%2F3%29r=100\"\r
\n" ); document.write( "\n" ); document.write( "\"r=100%283%2F10%29\"\r
\n" ); document.write( "\n" ); document.write( "\"highlight%28r=30%29\"------speed of slow train, 30 mph.
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