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: Understand what always true means here.
We are told only that \(x > 8\) and \(y > -4\). x and y can be any real numbers satisfying those two conditions, so a correct option must hold for every possible choice, not just some.
We test each option, either by finding a counterexample that breaks it, or by proving it algebraically from the two given conditions.

Step 2: Test option (A), \(xy < 0\).
Try \(x = 9\) (satisfies \(x>8\)) and \(y = 5\) (satisfies \(y>-4\)). Then \(xy = 45\), which is positive, not negative.
So option (A) is not always true.

Step 3: Test option (B), \(x^2 < -y\).
Since \(x > 8\), squaring gives \(x^2 > 64\).
Since \(y > -4\), we get \(-y < 4\).
So the claim needs a number bigger than 64 to be less than a number smaller than 4, which can never happen. Option (B) is never true, so it is certainly not always true.

Step 4: Test option (D), \(x > y\).
Try \(x = 9\) and \(y = 1000\), both allowed since \(9>8\) and \(1000>-4\). Here \(x = 9\) is smaller than \(y = 1000\), so \(x>y\) fails.
So option (D) is not always true either.

Step 5: Test option (C), \(-x < 2y\), algebraically.
From \(x > 8\), multiply both sides by \(-1\) (which flips the inequality):
\[ -x < -8 \]
From \(y > -4\), multiply both sides by \(2\) (multiplying by a positive number keeps the inequality direction):
\[ 2y > -8 \]
Now chain these two results together:
\[ -x < -8 < 2y \]
So \(-x < 2y\) always holds, no matter which valid \(x\) and \(y\) we pick, because \(-x\) is always below \(-8\) and \(2y\) is always above \(-8\).

Final Answer:
Only option (C), \(-x < 2y\), is guaranteed true for every \(x>8\) and \(y>-4\). \[ \boxed{-x < 2y} \]