Algebra.Com's Answer #853854 by ikleyn(53619)![\"\"](\"/images/stars/stars-13.gif\")  
You can put this solution on YOUR website!
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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( "less time find the 2 speeds.
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document.write( " The solution in the post by @mananth is incorrect.\r
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document.write( " It has an arithmetic error on the way, which leads to wrong answer.\r
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document.write( " I came to bring a correct solution.\r
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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) <<<---=== note it is DIFFERENT equation than that in the post by @mananth\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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document.write( "Solved correctly.\r
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