document.write( "Question 1153710: originally a rectangle was three times as long as it was wide. when 2 cm were subtracted from the length and 5 cm were added to the width, the resulting rectangle had an are of 90cm^2. what the dimensions of the new rectangle
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Algebra.Com's Answer #775988 by ikleyn(52781) $\"\"$ $\"About$

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document.write( "The original dimensions x cm wide and 3x cn long.\r\n" );
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document.write( "After changing it is (x+5) cm wide and (3x-2) cm long.\r\n" );
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document.write( "The equation\r\n" );
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document.write( "    (x+5)*(3x-2) = 90\r\n" );
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document.write( "You can solve it as a quadratic equation.\r\n" );
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document.write( "I will solve it mentally.\r\n" );
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document.write( "Multiply both sides by 3\r\n" );
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document.write( "    (3x+15)*(3x-2) = 270.\r\n" );
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document.write( "Find two integer factors of the number 270 that differ by 15-(-2) = 17.\r\n" );
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document.write( "Obviously, these factors are 27 and 10.\r\n" );
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document.write( "So, 3x-2 = 10;  3x = 10+2 = 12;  x = 12/3 = 4.\r\n" );
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document.write( "ANSWER:  The dimensions of the original rectangle are 4 cm (width) and 3*4 = 12 cm (length).\r\n" );
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