document.write( "Question 759806: A passenger train left Miami traveling north six hours before a freight train. The freight train traveled in the opposite direction going 35 mph slower then the passenger train for 12 hours after which time the trains were 2280 mi. apart. what was the passenger train's speed ? \n" ); document.write( "

Algebra.Com's Answer #462255 by Edwin McCravy(20067) $\"\"$ $\"About$

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A passenger train left Miami traveling north six hours before a freight train. The freight train traveled in the opposite direction going 35 mph slower then the passenger train for 12 hours after which time the trains were 2280 mi. apart. what was the passenger train's speed ?
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document.write( "Let x = the passenger train's speed.\r\n" );
document.write( "Then x-35 = the freight train's speed.\r\n" );
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document.write( "12 hours after the freight train left, which was 18 hours after the passenger\r\n" );
document.write( "train left, the passenger train was 18x miles north of the Miami station, and\r\n" );
document.write( " the freight train was 12(x-35) miles south of the Miami station.\r\n" );
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document.write( "Since 18 hours after the passenger train left going north is 12 hours after\r\n" );
document.write( "the freight train left Miami going south,  therefore,\r\n" );
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document.write( "           18x + 12(x-35) = 2280\r\n" );
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document.write( "Solve that and get 90 mph.\r\n" );
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document.write( "Edwin

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