If x and y are real numbers such that x \neq y, what is the value of \frac{x+y}{x-y}? (1) … — GMAT

If $x$ and $y$ are real numbers such that $x \neq y$ , what is the value of $\frac{x+y}{x-y}$ ?

(1) $x^2 + y^2 = 5xy$
(2) $x^2 - y^2 = 12$

A
Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.Answer
D
EACH statement ALONE is sufficient.
E
Statements (1) and (2) TOGETHER are NOT sufficient.

Both statements together are sufficient, but neither statement alone is sufficient.

Evaluating statement (1) reveals that

\left(\frac{x+y}{x-y}\right)^2 = \frac{7}{3}

, leading to two possible values:

\sqrt{\frac{7}{3}}

and

-\sqrt{\frac{7}{3}}

. Thus, statement (1) alone is insufficient. Statement (2) states that

(x+y)(x-y) = 12

, which shows that

(x+y)

and

(x-y)

have the same sign, meaning their quotient must be positive. Combining both statements eliminates the negative value, yielding the unique result

\sqrt{\frac{7}{3}}

. Therefore, both statements together are sufficient.

Analyze Statement (1) algebraically.

From

x^2 + y^2 = 5xy

, add

2xy

to both sides to obtain

(x+y)^2 = 7xy

, and subtract

2xy

from both sides to obtain

(x-y)^2 = 3xy

. Taking the ratio gives

\left(\frac{x+y}{x-y}\right)^2 = \frac{7xy}{3xy} = \frac{7}{3}

. Taking the square root yields

\frac{x+y}{x-y} = \pm \sqrt{\frac{7}{3}}

Since two distinct numerical values are possible, Statement (1) alone is NOT sufficient.

Analyze Statement (2) algebraically.

Statement (2) gives

x^2 - y^2 = 12

, which factors as

(x+y)(x-y) = 12

Knowing only the product of

(x+y)

and

(x-y)

does not provide enough information to determine the value of their quotient

\frac{x+y}{x-y}

. Statement (2) alone is NOT sufficient.

Combine Statement (1) and Statement (2).

Rewrite the target ratio using Statement (2):

\frac{x+y}{x-y} = \frac{(x+y)(x-y)}{(x-y)^2} = \frac{12}{(x-y)^2}

. Because

(x-y)^2 > 0

for

x \neq y

, the ratio must be strictly positive. Combining this with Statement (1), which requires

\frac{x+y}{x-y} = \pm \sqrt{\frac{7}{3}}

, uniquely isolates the positive solution

\frac{x+y}{x-y} = \sqrt{\frac{7}{3}}

Together, the statements determine a single, unique value for the target expression.

Algebraic Rephrasing and Sign Constraints in Data Sufficiency

Estimated Time:2m 0s

Question