If x is not equal to +or- y

If \(y>x\), then from the second statement we have \(x^2>y\) and from the first statement that \(x>y^2\). So, together we have \(x^2>y>x>y^2\).

In this case \(x^2>x\). Since \(x>0\), it means that \(x>1\)

But on the other hand \(y>y^2\) and \(1>y\).

So, we have \(x>1>y\), that contradicts \(y>x\).

Therefore, \(y-x<0\) and \(x+y>0\), since \(x\)and \(y\) is positive.

Thanks Smaryaga.I was confused initially with the formation of combined statement "x^2>y>x>y^2"
Though you have just formatted the explanation stated previously but this time I have understand the idea behind it.

Sometimes, it is difficult to understand what is not understandable for other person))) Next time may be better to ask the question more specific)))

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

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

\(\frac{(x-y)}{(x+y)}>\frac{(x+y)}{(y-x)}\)

add 1 to each side of the fraction

\(\frac{(x-y)}{(x+y)}\) \(+ 1 >\) \(\frac{(x+y)}{(y-x)} +1\)

thus we have IS \(\frac{x}{x+y}\) \(>\)\(\frac{y}{y-x}\) ??

st.1
x>y^2
as y^2 is always greater than zero. thus x is greater than zero.

for y=0.2 and x=0.1 then answer to the original question is no. whereas for y=2 and x=3 answer is yes. hence st.1 alone is not sufficient

st.2

x^2>y>0

for x=2, and y=3 answer to the original question is yes. whereas for x=3 and y=2 answer to the original question is no. hence st.2 alone is not sufficient.

combining st.1 and st.2

we know that x>y. hence answer to the original question is yes. thus answer is C.

If we just simplify (x-y)/(x+y)>(x+y)/(y-x)

We will get (x-y)^2 > - (x+y)^2.

On GMAT, we have only real numbers and x != + or - y. Won't that equation always hold true?

Where am I getting wrong?

(x-y)/(x+y) > -(x+y)/(x-y): we cannot cross-multiply here because we don't know the signs of the denominators: (x+y) and (x-y). Recall that we should flip the sign of an inequality if we multiply it by negative value.

If (x+y) and (x-y) have the same sign, for example if both are positive, then when cross-multiplying we get (x-y)^2 > -(x+y)^2. But if (x+y) and (x-y) have the opposite signs, for example if (x+y) is positive and (x-y) is negative, the when cross-multiplying we should flip the sign of the inequality and we get (x-y)^2 < -(x+y)^2.

Never multiply (or reduce) an inequality by a variable (or the expression with a variable) if you don't know its sign.


Hope it helps.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If x is not equal ±y, is
(x−y)(x+y) >(x+y)(y−x) ?

1)x>y 2
2)x 2 >y>0

We get (x-y)/(x+y)+(x+y)/(x-y)>0? --> [(x-y)^2+(x+y)^2]/(x^2-y^2)>0? --> 2(x^2+y^2)/(x^2-y^2)>0? and 1/(x^2-y^2>0? as x^2+y^2>0 if we modify the question.
There are 2 variables (x,y) and 2 equations are given by the 2 conditions, so there is high chance (C) will become the answer, and (C) is actually the answer.

For cases where we need 2 more equations, such as original conditions with “2 variables”, or “3 variables and 1 equation”, or “4 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 70% chance that C is the answer, while E has 25% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since C is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, D or E.

oh man...i mistakenly assumed that x and y are integers, and selected A...

stem can be manipulated down to is 4xy / x^2-y^2 >0

from the 2 statments

x,y + ve , x>y^2 and thus x^2> y^2 .... suff
C

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