document.write( "Question 1031863: frank went 16 miles at one speed and then caame back going 4mph faster. if the return trip took 40mins less time find the 2 speeds. \n" ); document.write( "

Algebra.Com's Answer #853855 by n2(54) $\"\"$ $\"About$

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\n" ); document.write( "frank went 16 miles at one speed and then came back going 4mph faster. if the return trip took 40 mins
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document.write( "Let x be the rate going to there, in miles per hour.\r\n" );
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document.write( "Then the rate going back is (x+4) miler per hour.\r\n" );
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document.write( "The time going at one speed is    hours.\r\n" );
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document.write( "The time going back is    hours.\r\n" );
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document.write( "The time equation is\r\n" );
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document.write( "     -  = .\r\n" );
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document.write( "To solve, multiply both sides by LCD  3x*(x+4).  You will get\r\n" );
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document.write( "    48(x+4) - 48x = 2x(x+4)     \r\n" );
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document.write( "    192 = 2x(x+4)\r\n" );
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document.write( "     96 = x(x+4)\r\n" );
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document.write( "    x^2 + 4x - 96 = 0\r\n" );
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document.write( "    (x+12)*(x-8) = 0\r\n" );
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document.write( "The roots are -12 and 8.  We reject negative root and accept the positive one x = 8.\r\n" );
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document.write( "ANSWER.  The speeds are 8 mph (going to there) and 12 mph (going back).\r\n" );
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document.write( "CHECK.  The time going to there is   = 2 hours.\r\n" );
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document.write( "        The time going back is   =  =  = 1 hours.\r\n" );
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document.write( "        The difference is    of an hour,  which is PRECISELY CORRECT.\r\n" );
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