document.write( "Question 1169508: Help me solve this, please!
\n" ); document.write( "The length of a rectangle is one more than the width. If the dimensions are both decreased by 2 units, the ares of the new rectangle is 30 sq. units less than the area of the original rectangle. Find the area of the original rectangle. \n" ); document.write( "

Algebra.Com's Answer #794260 by ikleyn(52915) $\"\"$ $\"About$

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document.write( "Let x be the width of the original rectangle.\r\n" );
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document.write( "Then its length is (x+1), according to the condition,\r\n" );
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document.write( "    and the area is  x*(x+1).\r\n" );
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document.write( "The decreased dimensions are  (x-2) and ((x+1)-2) = (x-1),\r\n" );
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document.write( "so the decreased area is  (x-2)*(x-1).\r\n" );
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document.write( "From the condition,\r\n" );
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document.write( "    (x-2)*(x-1) = x*(x+1) - 30.\r\n" );
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document.write( "Simplify\r\n" );
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document.write( "    x^2 - 3x + 2 = x^2 + x - 30\r\n" );
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document.write( "        30 + 2   = x + 3x\r\n" );
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document.write( "          32     = 4x\r\n" );
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document.write( "          x      = 32/4 = 8.\r\n" );
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document.write( "ANSWER.  The original dimensions are  8 units (the width)  and  8+1 = 9 units (the length).\r\n" );
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