MackyCee wrote:

From OG11...

Q139) if "x" is not equal to "-y", is x-y / x+y > 1?

(1) X > 0

(2) Y < 0

We know that we cannot multiply an inequality by an unknown if we don't know whether the unknown is a positive value or a negative value. The inequality behaves differently in the two cases. If the unknown is positive, the inequality stays as it is. If the unknown is negative, the inequality flips sign.

Therefore, x-y > x+y is not an option.

Next, '>1' is much more complicated that '>0' where we just need to consider whether the variables are positive or negative. Therefore, bring whatever is on the right hand side to the left hand side.

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

Is \(\frac{(x - y)}{(x + y)} - 1 > 0\)

Is \(\frac{-2y}{(x + y)} > 0\)

For this expression to be positive, y/(x+y) should be negative. Therefore, either 'y should be negative and (x+y) should be positive' or 'y should be positive and (x+y) should be negative'.

Statement 1: X > 0

No info about y so not sufficient.

Statement 2: Y < 0

No info about x so not sufficient.

Both together, we know that y is negative. We now need to know the sign of (x+y). Just knowing that x is positive doesn't tell us the sign of (x+y) because we don't know which one out of x and y has greater absolute value. If absolute value of x is greater than that of y, x+y is positive. If absolute value of x is less than that of y, x+y is negative. We do not know the sign of x+y so we still cannot say whether \(\frac{-2y}{(x + y)}\) is positive. Not sufficient.

Answer E.

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