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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 + 8A. 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 > 4So 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 > 4So 5(-x² – 1) < 4(-x² – 1)

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