If x and y are non-zero rational numbers such that \frac{x}{y} 0, which of the following statements must be true? Select all that apply.

If x and y are non-zero rational numbers such that \frac{x}{y} < -\frac{1}{2} and x + y > … — GRE General Test

If $x$ and $y$ are non-zero rational numbers such that $\frac{x}{y} < -\frac{1}{2}$ and $x + y > 0$ , which of the following statements must be true? Select all that apply.

$xy < 0$ Answer
$\frac{1}{x} + \frac{1}{y} < 0$ Answer
C
$x > y$
D
$|x| > |y|$
$x^2 + y^2 > (x+y)^2$ Answer

Answer

The statements

xy < 0

\frac{1}{x} + \frac{1}{y} < 0

, and

x^2 + y^2 > (x+y)^2

must be true.

Because

\frac{x}{y} < -\frac{1}{2}

x

and

y

must carry opposite algebraic signs, establishing that

xy < 0

. Combining the reciprocals into

\frac{x+y}{xy}

places a positive numerator over a negative denominator, ensuring the sum of reciprocals is strictly negative. Expanding

(x+y)^2 = x^2 + y^2 + 2xy

shows that adding the negative term

2xy

makes

(x+y)^2 < x^2 + y^2

Step-by-Step Solution

Determine the sign of the product

xy

xy < 0

The quotient of two non-zero real numbers is negative if and only if they have opposite signs. Since

\frac{x}{y} < -\frac{1}{2} < 0

x

and

y

must have opposite signs, so their product is negative.

Evaluate the sum of reciprocals

\frac{1}{x} + \frac{1}{y}

\frac{1}{x} + \frac{1}{y} < 0

Finding a common denominator gives

\frac{1}{x} + \frac{1}{y} = \frac{x+y}{xy}

. Given

x+y > 0

(positive) and

xy < 0

(negative), dividing a positive number by a negative number yields a negative result.

Compare

x^2 + y^2

with

(x+y)^2

x^2 + y^2 > (x+y)^2

Using algebraic expansion,

(x+y)^2 = x^2 + y^2 + 2xy

. Because

xy < 0

2xy

is negative. Subtracting a positive value (or adding a negative value) to

x^2 + y^2

results in a smaller quantity.

Test counterexamples for

x > y

and

|x| > |y|

Neither statement is required to be true.

Let

x = -2

and

y = 3

. Both are rational numbers. Check conditions:

\frac{-2}{3} < -\frac{1}{2}

holds, and

-2 + 3 = 1 > 0

holds. For this counterexample,

x = -2 < 3 = y

and

|-2| = 2 < 3 = |y|

Key Concept

Signs and inequalities of rational numbers and reciprocal operations

Question

Answer

Step-by-Step Solution

Key Concept

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