document.write( "Question 1166715: Two rectangles have the same area of 24cm^2. The second rectangle is 4cm shorter and 1cm wider than the first. What is the length and the breadth of the first rectangle?
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Algebra.Com's Answer #791245 by ikleyn(52781) $\"\"$ $\"About$

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document.write( "Let x and y be the dimensions of the first rectangle.\r\n" );
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document.write( "We have  xy = 24 for its area.\r\n" );
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document.write( "The dimensions of the second rectangle are (x-4) and (y+1)  with the equation for the area\r\n" );
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document.write( "    (x-4)*(y+1) = 24,  or\r\n" );
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document.write( "    xy + x - 4y - 4 = 24.\r\n" );
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document.write( "Replacing here xy by 24, based on the very first equation, we get\r\n" );
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document.write( "    24 + x - 4y - 4 = 24, \r\n" );
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document.write( "or, after collecting/canceling common terms\r\n" );
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document.write( "    x - 4y = 4.    (*)\r\n" );
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document.write( "So, we have now two equations\r\n" );
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document.write( "    xy = 24      (1)\r\n" );
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document.write( "    x = 4 + 4y.  (2)\r\n" );
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document.write( "By substituting (2) to (1), you get a quadratic equation\r\n" );
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document.write( "    (4+4y)*y = 24\r\n" );
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document.write( "    (1+y)*y = 6\r\n" );
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document.write( "At this point, you can solve it as a quadratic equation\r\n" );
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document.write( "or GUESS the solution mentally  y = 2.\r\n" );
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document.write( "ANSWER.  The dimensions of the first rectangle are  2 cm (the width)  and  24/2 = 12 cm (the length).\r\n" );
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