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GRE Algebra (Quantitative Comparison) http://www.800score.com/forum/viewtopic.php?f=12&t=3283 
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Author:  questioner [ Sat Mar 31, 2012 9:22 am ] 
Post subject:  GRE Algebra (Quantitative Comparison) 
Quantity A: (5x² – 20)/(x – 2) Quantity B: 4x + 8 A. Quantity A is greater. B. Quantity B is greater. C. The two quantities are equal. D. The relationship cannot be determined from the information given. The correct answer is (D). To solve this, you must simplify Quantity A. Factor the common 5 in the numerator to make 5(x² – 4)/(x – 2) and then factor the difference of squares in the numerator: 5(x – 2)(x + 2)/(x – 2). Now you can cancel out the (x – 2) from both the numerator and denominator to get 5(x + 2). Factor 4 in Quantity B. It becomes 4(x + 2). Now we see that the both quantities have the same factor, (x + 2), which becomes 0 if x = 2. In this case quantities will be equal. However, if (x + 2) is positive, then quantity A will be greater because 5 > 4. We already have two different possibilities, so the relationship cannot be determined from the information given. The correct answer is D.  Can't you divide both quantities by (x + 2) so that A = 5 and B = 8? 
Author:  Gennadiy [ Sat Mar 31, 2012 9:32 am ] 
Post subject:  Re: t.1, s.2, qt.7: algebra, manipulating expressions 
Quote: Can't you divide both quantities by (x + 2) ... ? No, we cannot, because we do NOT know if it is positive, negative or even 0! You cannot divide by 0. If (x + 2) = 0, then the quantities are equal. If (x + 2) < 0, then by dividing the both parts of the equation by a negative number, you change the sign to an opposite one. Since 5 > 4, then the original sign must be 5(x + 2) < 4(x + 2). If (x + 2) > 0, then by dividing the both parts of the equation by a positive number, you do NOT change the sign. Since 5 > 4, then the original sign must be 5(x + 2) > 4(x + 2). You may try to plug in some values of x less than 2 and greater than 2 to get the full understanding. IF there had been 5(x² + 1) and 4(x² + 1), then we could have divided the inequality by (x² + 1). (x² + 1) is always positive. The sign would NOT have changed. 5(x² + 1) ? 4(x² + 1) 5 > 4 So 5(x² + 1) > 4(x² + 1) IF there had been 5(x² – 1) and 4(x² – 1), then we either could have factored (1) first, or we could have divided the inequality by (x² – 1) right away. (x² – 1) is always negative, so the sign would have changed. 5(x² – 1) ? 4(x² – 1) 5 > 4 So 5(x² – 1) < 4(x² – 1) 
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