document.write( "Question 465605: The length of a rectangle is 15 cm longer than the width. when its width is increased by 3 cm and its length decreased by 5 cm. the area of the new rectangle is 20cm^2 bigger than the original. find the original dimension of the rectangle. \n" ); document.write( "

Algebra.Com's Answer #319041 by ewatrrr(24785) $\"\"$ $\"About$

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document.write( "Hi,
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document.write( "Let x and (x+15cm) represent the width and length of the 1st rectangle and
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document.write( "(x+3), (x+10) the width and length of the 2nd rectangle
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document.write( "Question states***
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document.write( "(x+3)(x+10) = x(x+15) + 20cm^2
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document.write( "Solving for x
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document.write( "x^2 + 13x + 30 = x^2 + 15x + 20
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document.write( "             10 = 2x
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document.write( "            x = 5cm, width of the original.  Length is 20cm.\r
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document.write( "CHECKING our Answer*** 2nd rectangle: width is 8cm and Length is 15cm
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document.write( "    120cm^2 = 100cm^2 + 20cm^2 \n" );
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