document.write( "Question 947456: I've been working on this problem for hours and still can't seem to find a solution, could you please help if possible?\r
\n" ); document.write( "\n" ); document.write( "A person drives to a destination at a rate of thirty mph and returns over the same route at fifty mph. How far is the destination if the time returning is one hour less than the time going? \n" ); document.write( "

Algebra.Com's Answer #578141 by MathTherapy(10552) $\"\"$ $\"About$

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\n" ); document.write( "I've been working on this problem for hours and still can't seem to find a solution, could you please help if possible?\r
\n" ); document.write( "\n" ); document.write( "A person drives to a destination at a rate of thirty mph and returns over the same route at fifty mph. How far is the destination if the time returning is one hour less than the time going?
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document.write( "Let the distance be D
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document.write( "Then outbound time = 
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document.write( "Inbound time = 
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document.write( "Therefore, 
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document.write( "5D – 150 = 3D ------ Multiplying by LCD, 150
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document.write( "5D – 3D = 150
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document.write( "2D = 150
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document.write( "D, or distance = , or  mph \n" );
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