document.write( "Question 72629: Solve and show work.
\n" ); document.write( "At 1:00 p.m., a car leaves a city and travels north at a rate of 55 mi/h. An hour later, a second car leaves the city and travels south at a rate of 60 mi/hr. At what time will the two cars be 285 miles apart? \n" ); document.write( "
Algebra.Com's Answer #51933 by Edwin McCravy(20055)\"\" \"About 
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document.write( "At 1:00 p.m., a car leaves a city and travels north at \r\n" );
document.write( "a rate of 55 mi/h. An hour later, a second car leaves\r\n" );
document.write( "the city and travels south at a rate of 60 mi/hr. At \r\n" );
document.write( "what time will the two cars be 285 miles apart?\r\n" );
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document.write( "Let the time when they meet be t hours after 1:00 p.m., \r\n" );
document.write( "which is the total time the first car traveled.\r\n" );
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document.write( " FIRST        SECOND         THEIR\r\n" );
document.write( " CAR'S    +   CAR'S    =   DISTANCE\r\n" );
document.write( "DISTANCE     DISTANCE        APART\r\n" );
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document.write( "We use DISTANCE = RATE × TIME\r\n" );
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document.write( "The first car's RATE is 55 and its TIME is t hours, so\r\n" );
document.write( "its distance is 55t\r\n" );
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document.write( "The second car's RATE is 60 and its TIME is 1 hour less\r\n" );
document.write( "than the first car's, or t-1. So its distance is 60(t-1)\r\n" );
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document.write( " FIRST        SECOND         THEIR\r\n" );
document.write( " CAR'S    +   CAR'S    =   DISTANCE\r\n" );
document.write( "DISTANCE     DISTANCE        APART\r\n" );
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document.write( "  55t     +  60(t - 1) =      285\r\n" );
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document.write( "Solve that and get t = 3 hours after 1:00 p.m., or 4:00 p.m.\r\n" );
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document.write( "Edwin
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