Question

Any non-zero real numbers \(x, y\) such that \(y ≠ 3\) and \(\frac{x}{y}<\frac{x+3}{y-3}\), will satisfy the condition

• A. \(\frac{x}{y}<\frac{y}{x}\)
• B. If \(y >10\) , then \(− x > y\)
• C. If \(x < 0\) , then \(− x < y\)
• D. If \(y < 0\) , then \(− x < y\)

Answer 1

The Correct Option is D

Given:
\(\frac{x}{y} < \frac{x+3}{y-3}\) 

This inequality can be rewritten as:

\[ \frac{x}{y} - \frac{x+3}{y-3} < 0 \]

Now, let's combine the two fractions:

\[ \Rightarrow \frac{x(y-3) - y(x+3)}{y(y-3)} < 0 \]

Expanding the numerator:

\[ \Rightarrow \frac{xy - 3x - xy - 3y}{y(y - 3)} < 0 \] \[ \Rightarrow \frac{-3(x + y)}{y(y - 3)} < 0 \]

Multiplying both sides by -1 (to make the numerator positive):

\[ \Rightarrow \frac{3(x + y)}{y(y - 3)} > 0 \]

Now, let's analyze the inequality \( \frac{3(x + y)}{y(y - 3)} > 0 \).

For this inequality to be true, we need to consider the conditions for both the numerator and the denominator. We know that:

If \( y < 0 \), then \( y(y - 3) > 0 \) (since the product of two negative numbers is positive). Thus, for the inequality \( \frac{3(x + y)}{y(y - 3)} > 0 \) to hold true, we must have:

\[ x + y > 0 \]

This implies:

\[ x > -y \]

Additionally, since \( x \) and \( y \) are related by their absolute values, we conclude that:

\[ |x| > |y| \]

Therefore, the final conclusion is:

\[ -x < y \]

Conclusion: Therefore, the correct option is (D): "If \( y < 0 \), then \( -x < y \)."