Question

If \(x > 8\) and \(y > -4\), then which one of the following is always true?

• A. \(xy < 0\)
• B. \(x^2 < -y\)
• C. \(-x < 2y\)
• D. \(x > y\)

Hint:

Turn the two given conditions into -x less than -8 and 2y more than -8, then chain them to see which option they force to be true.

Answer 1

The Correct Option is C

Step 1: Picture the conditions on a number line.
x is any number strictly to the right of 8, so x can be 8.0001, 100, or a million. y is any number strictly to the right of -4, so y can be -3.999, 0, or a million as well. Since neither variable has an upper bound, any option involving a fixed relationship between the sizes of x and y needs to be checked at extreme values, not just typical ones.

Step 2: Break options (A) and (D) using large values.
For (A), $xy < 0$ needs one of x, y to be negative. But x is always positive here (since $x > 8 > 0$), and y is free to be positive too (say y = 1), which makes xy positive. So (A) fails for that choice.
For (D), $x > y$ needs y to stay smaller than x always. But since y has no upper limit, picking y = 1,000,000 with x = 9 breaks $x > y$ right away.

Step 3: Break option (B) using the lower bounds.
The smallest $x^2$ can get close to is $8^2 = 64$ (approached but never reached, since x is strictly greater than 8), and the largest $-y$ can get close to is $4$ (as y approaches $-4$ from above). Since $x^2$ is always above 64 and $-y$ is always below 4, the required inequality $x^2 < -y$ can never be satisfied.

Step 4: Confirm option (C) using the extreme boundary values.
Push x to its smallest allowed value, just above 8, and y to its smallest allowed value, just above -4. Even in this tightest case, $-x$ sits just below $-8$, and $2y$ sits just above $-8$, keeping $-x < 2y$ true.
Now push x and y to be very large instead: $-x$ becomes a huge negative number, while $2y$ becomes a huge positive number, so $-x < 2y$ is even more clearly true.
Since the inequality survives both the tightest boundary case and the most extreme case, it holds everywhere in between as well.

Step 5: Conclude.
Only $-x < 2y$, option (C), survives every test, so it is the one that is always true.
\[ \boxed{-x < 2y \text{ is always true}} \]