If x

Question

If x < -1 , is \frac{x+y}{x+1}<\frac{x-y}{x-1} ? (1) x + y > 0 (2) y > 0

yashikaaggarwal · Answer

Given: X < -1
To find: X+Y/X+1 < X-Y/X-1

Solution:
(1) X+Y > 0
Therefore Y is greater than 2 (since X lowest value is -2)
Let X = -1.1 and Y = 2.1 (lowest possible value)
X+Y/X+1 < X-Y/X-1
-1.1+2.1/-1.1+1 < -1.1-2.1/-1.1-1
1/-0.1 < -3.2/-2.1
-10 < 1.52
(Sufficient)

(2) Y > 0
Case 1:
X = -1.1 and Y = 0.1
X+Y/X+1 < X-Y/X-1
-1.1+0.1/-1.1+1 < -1.1-0.1/-1.1-1
-1/-0.1 < -1.2/-2.1
10 > 0.57 (Wrong)

Case 2: Let X = -1.1 and Y = 2.1
X+Y/X+1 < X-Y/X-1
-1.1+2.1/-1.1+1 < -1.1-2.1/-1.1-1
1/-0.1 < -3.2/-2.1
-10 < 1.52 (Correct)
(Insufficient)

Answer is A

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BrentGMATPrepNow · Answer

I should mention that y can be less than 2. 
For example, x = -1.01 and y = 1.02 would meet the condition that x+y > 0

BrentGMATPrepNow · Answer

Target question : Is \frac{x+y}{x+1}<\frac{x-y}{x-1} ? This is a great candidate for rephrasing the target question . Given: \frac{x+y}{x+1}<\frac{x-y}{x-1} Multiply both sides of the inequality by x+1 to get: x+y>\frac{(x-y)(x+1)}{x-1} Important: since x < -1 , we know that x+1 is NEGATIVE. So when we multiply by x+1 , we must REVERSE the direction of the inequality symbol. Multiply both sides of the inequality by x-1 to get: (x+y)(x-1)<(x-y)(x+1) Note: Since x-1 is NEGATIVE, we REVERSED the direction of the inequality symbol. Expand both sides to get: x^2 - x + xy - y1 Note: Since 2x is NEGATIVE, I REVERSED the direction of the inequality symbol after dividing by 2x . REPHRASED target question : Is y>1 ? Aside: the video below has tips on rephrasing the target question At this point, the statements will be very easy to analyze. Statement 1 : x + y > 0 We already know that x < -1 , which I'll rewrite as -1 > x Since the inequality symbols in -1 > x and x + y > 0 are facing the same direction, we can add the inequalities to get: x + y - 1 > x Subtract x from both sides: y - 1 > 0 Add 1 to both sides: y > 1 So, the answer to the REPHRASED target question is YES, y is greater than 1 Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT Statement 2 : y > 0 There are several values of y that satisfy statement 2. Here are two: Case a : y = 3, in which case, the answer to the REPHRASED target question is YES, y is greater than 1 Case b : y = 0.5, in which case, the answer to the REPHRASED target question is NO, y is not greater than 1 Since we can’t answer the target question with certainty, statement 2 is NOT SUFFICIENT Answer: A Cheers, Brent VIDEO ON REPHRASING THE TARGET QUESTION: https://www.youtube.com/watch?v=MyG1K3ee69w

If x < -1, is x+y/x+1 < x-y/x-1?

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