If x is not equal to |y|, is (x - y)/(x + y) (x + y)/(x - y) ?

Question

If x is not equal to |y|, is  \frac{x-y}{x+y} > \frac{x+y}{x-y} ?

(1)  x > y^2

(2)  x^2>y>0

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PyjamaScientist · Answer

Is    \frac{x-y}{x+y} > \frac{x+y}{x-y} ?

(I)   x>y^2 
This gives,  x>y   or  x-y>0 . But what about  x+y ? What if  y=-4  and  x=4 . That would give  x+y=0  And the L.H.S would become not defined. So, this alone is  insufficient .

(II)   x^2>y>0 
This can lead to three possibilities,
 Case (A)   x<y (x=2,y=3,x^2=4) 
In this case, is  \frac{x-y}{x+y} > \frac{x+y}{x-y}  =  \frac{2-3}{2+3} > \frac{2+3}{2-3}  =  \frac{-1}{5} > \frac{-5}{1} . So, Yes!  \frac{x-y}{x+y} > \frac{x+y}{x-y}    holds .

Case (B)   x=y (x=2,y=2) 
For this x-y = 2-2= 0. This makes R.H.S not defined. Not possible.

Case (C)  x>y (x=3,y=2) 
 \frac{x-y}{x+y} > \frac{x+y}{x-y}  =  \frac{3-2}{3+2} > \frac{3+2}{3-2}  =  \frac{1}{5} > \frac{5}{1} ? No!  \frac{x-y}{x+y} > \frac{x+y}{x-y}    does not hold .
Thus,  insufficient , since cases (A) and (C) give contradictory answers.

(I) + (II)  =  x>y 
In that case (C) gives a definite unique answer=No. Thus, both are sufficient together.
So, answer is option  (C) .

cjhyou700 · Answer

The question already tell us "x is not equal to |y|". I don't understand why we consider the situation y=-4, and x=4.

cyrwr1 · Answer

Stmt (1) makes you determine x is positive
Stmt (2) makes you determine y is positive

with both statements, you have both x and y (+) and x is larger than y, regardless how you put the four variables on a number line

the fractions will be smaller positive divided by a (+) vs. (+) divided by smaller positive

left side is ALWAYS smaller

cjhyou700 · Answer

cyrwr1 
Hello. Thanks for your reply. I get the answer now. I think my first approach was wrong, so I screwed up later steps.

ahamd · Answer

If we break down the given equation,It comes down to if 4xy<0.Hence X & should be of opposite signs.

If x is not equal to |y|, is (x - y)/(x + y) > (x + y)/(x - y) ?

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