Let x and y be non-zero integers such that -6 \le x \le 6 and -6 \le y \le 6. If x^3 y < 0… — GMAT

Let $x$ and $y$ be non-zero integers such that $-6 \le x \le 6$ and $-6 \le y \le 6$ . If $x^3 y < 0$ and $\frac{x}{y^2} < -\frac{1}{4}$ , what is the minimum possible value of $x - y$ ?

Answer: -10

Answer

The minimum possible value of

x - y

-10

Analyzing the signs shows x must be negative and y must be positive. Multiplying x/y² < -1/4 by 4y² gives y² < -4x. Testing x = -6 gives y² < 24, so the largest positive integer y is 4. The expression x - y reaches its minimum value of -6 - 4 = -10.

Step-by-Step Solution

Determine the signs of variables x and y

x < 0 and y > 0

Since y² > 0 for any non-zero integer y, x/y² < -1/4 requires x < 0. Furthermore, x³y < 0 requires x³ and y to have opposite signs; since x < 0 implies x³ < 0, y must be positive.

Transform the inequality without altering its sign direction

y² < -4x

Multiplying x/y² < -1/4 by 4y² > 0 yields 4x < -y², which rearranges to y² < -4x.

Evaluate candidate integer pairs to minimize x - y

The minimum value of x - y is -10 when x = -6 and y = 4

To minimize x - y, select the most negative integer x and the largest allowable positive integer y. Setting x = -6 gives y² < 24, making max integer y = 4. Therefore, x - y = -6 - 4 = -10.

Key Concept

Deducing signs from products and quotients, and manipulating inequalities involving non-zero variables.

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