Systematic Inequality Solving

j41981 · Posted: Wed Jul 19, 2006 11:23 pm Post subject: Systematic Inequality Solving

Hi all,
I often come across a DS problem involving inequalities, and though I try to systematically solve it, I either find myself resorting to plugging in, or getting the problem wrong if I don't.

Here's an example: x does not equal -y, is (x-y)/(x+y)>1?

(1) x>0
(2) y<0

Seeing this setup, I would immediately try and simplify the expression. I would group the problem into two scenarios: either (x+y) is (a) positive or (b) negative.

(a) If (x+y) is positive, then you can multiply both sides by that term without flipping the sign, hence the question is asking: is (x-y)>(x+y)? This simplifies further to: is -y>y? I guess this would imply that y must be negative.

(b) If (x+y) is negative, then you must flip the sign after multiplying. The question becomes: is (x-y)<(x+y)? This simplifies to: is -y<y? I guess this would imply that y must be positive.

Statement (1) seems insufficient, since x cancels out in both scenarios. Statement (2) would seem to address both scenarios, and yield an answer.

However, checking by plugging in (as well as looking at the answer) indicates that the answer is E. I've decided maybe I'm forgetting something fundamental about simplifying algebraic inequalities, and so my systematic attempts (as opposed to plugging in) are faulty. Or, maybe Ive set it up correctly but my logic is faulty. Can anyone let me know how they would approach an inequalities problem where one might create "scenarios" based on an algebraic inequality?

Thanks, your help is appreciated.

jayofbay · Posted: Thu Jul 20, 2006 1:44 pm Post subject:

You are forgetting the case where x=1 and y=-1 the answer at that point is not solvable.

rehanaj · Posted: Thu Jul 20, 2006 2:00 pm Post subject:

You are pretty much on the right track. I will start off where you (incorrectly) left off. What I have added is included in square [...] brackets:

(a) If (x+y) is positive, then you can multiply both sides by that term without flipping the sign, hence the question is asking: is (x-y)>(x+y)? This simplifies further to: is -y>y? I guess this would imply that y must be negative.

[ So y < 0, and we have assumed that x + y > 0. This is possible if and only if x > 0. Combining the two, we need x > 0 AND y < 0. Statment 1 is thus not sufficient, because it includes TWO scenarios, one TRUE and one FALSE
x > 0, y < 0 True
x > 0, y < 0 False ]

(b) If (x+y) is negative, then you must flip the sign after multiplying. The question becomes: is (x-y)<(x+y)? This simplifies to: is -y<y? I guess this would imply that y must be positive.

[ So y > 0, and we have assumed that x - y > 0. This is possible if and only if x < 0. Combining the two, we need x < 0 AND y > 0. Statment 2 is thus not sufficient, because it includes TWO FALSE scenarios
x > 0, y < 0 False
x < 0, y < 0 False ]

[We have thus proved that neither statement taken alone is sufficient].

[Now, take the 2 statements together, x > 0 and y < 0 and what you are given, i.e. x .ne. y]

[We have shown that when x > 0 and y < 0, and x + y > 0, i.e. |x| > |y|, then given expression is True.

We have also shown that when x < 0 and y > 0 and x + y < 0, i.e. |x| > |y|, then the given expression is true. But this second part is irrelevant.

Together, thus, given the two conditions stipulated by statements (1) and (2) we can definitively say that the expression is TRUE only if x| > |y|]

To summarize

when x > 0 and y < 0, and |x| > |y|, then given expression is True.

when x > 0 and y < 0, and |x| < |y|, then given expression is False.

Hence 2 statements taken together are therefore not sufficient.]

Rehana

gmat_enthus · Posted: Fri Nov 24, 2006 1:59 pm Post subject: Re: Systematic Inequality Solving

Mark Dabral · Posted: Tue Nov 28, 2006 4:28 am Post subject: My take on inequalities

Please see the attached explanation. This is my first try at a free hand explanation using a stylus and a writing software. I hope my handwriting is legible.

Cheers,
Mark