document.write( "Question 1127986: two planes leave an airport flying at the same rate. If the first plane flies 1.5 hours longer than the the second plane and travels 2700 miles while the second plane travels only 2025 miles, for how long was each plane flying?
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Algebra.Com's Answer #744467 by ikleyn(52803) $\"\"$ $\"About$

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document.write( "From the condition, you have these two equations\r\n" );
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document.write( "v*(t+1.5) = 2700     (1)   (\"the first plane flies 1.5 hours longer than the the second plane and travels 2700 miles\")\r\n" );
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document.write( "v*t       = 2025     (2)   (\"the second travels 2025 miles at the same rate\")\r\n" );
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document.write( "where v is their common rate and t is flying time for the second plane.\r\n" );
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document.write( "Divide equation (1) by equation (2) (both sides.  You will get\r\n" );
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document.write( " =  = \r\n" );
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document.write( "3*(t+1.5) = 4t\r\n" );
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document.write( "3t + 4.5 = 4t\r\n" );
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document.write( "t = 4.5 hours.\r\n" );
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document.write( "Answer.  The second plane was flying for 4.5 hours;  the first plane was flying for  4.5 + 1.5 = 6 hours.\r\n" );
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document.write( "     The ratio of distances is   = ;  hence, the ratio of times is  , too \r\n" );
document.write( "     (since the rate is the same for both planes).\r\n" );
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document.write( "    Hence, first plane traveled 4x hours, while the second plane traveled 3x hours.\r\n" );
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document.write( "    We are given that  4x - 3x  is  1.5 hours;  hence  x = 1.5 hours.\r\n" );
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document.write( "    Then the first plane was flying  4*1.5= 6 hours, while the second plane was flying 3*1.5 = 4.5 hours.\r\n" );
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document.write( "    You got the same answer.\r\n" );
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