document.write( "Question 1153922: Hi\r
\n" ); document.write( "\n" ); document.write( "A plane took off 1 hour later for its destination 3000km away. In order to reach the destination on time the pilot increased speed by 500km per hour. \r
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Algebra.Com's Answer #776247 by ikleyn(52781) $\"\"$ $\"About$

You can put this solution on YOUR website!
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document.write( "Let x be the new speed, in km/h.\r\n" );
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document.write( "Then the regular speed is (x-500) km/h.\r\n" );
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document.write( "The regular flight time is   hours.\r\n" );
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document.write( "The new time flight is       hours.\r\n" );
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document.write( "The difference is one hour (1 hour), which gives you THIS \"time equation\"\r\n" );
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document.write( "     -  = 1   hour.\r\n" );
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document.write( "To solve the equation, multiply both sides by  x*(x-500).  You will get\r\n" );
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document.write( "    3000x - 3000*(x-500) = x*(x-500)\r\n" );
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document.write( "    300*500 = x*(x-500)\r\n" );
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document.write( "    x^2 - 500x - 1500000 = 0\r\n" );
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document.write( "     =  =  = .\r\n" );
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document.write( "Only positive root makes sense  x =  = 1500.\r\n" );
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document.write( "ANSWER.  The new speed is 1500 km/h.\r\n" );
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document.write( "CHECK.   The regular speed is then 500 km/h lover, i.e.  1000 km/h.\r\n" );
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document.write( "         Check the \"time equation\"   -  = 3 - 2 = 1 hour.   ! Precisely correct !\r\n" );
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