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 can be expressed as :

\(\frac{x}{y}-\frac{x+3}{y-3}<0\)
Now,
⇒ \(\frac{x(y-3)-y(x+3)}{y(y-3)}<0\)

⇒ \(\frac{xy-3x-xy-3y}{y(y-3)}<0\)

⇒ \(\frac{-3(x+y)}{y(y-3)}<0\)
 

⇒ \(\frac{3(x+y)}{y(y-3)}>0\)
From the above inequality , we can say that :
when y < 0 ⇒ y(y - 3) > 0.
So, to satisfy the above given equation \(\frac{3(x+y)}{y(y-3)}>0\),
(x + y) must be greater than zero.
Therefore, x > 0 and |x| > |y|
So, the magnitude of x is greater than the magnitude of y.
Therefore, x > y and |x| > |y| ⇒  -x < y

So, the correct option is (D) : If \(y < 0\) , then \(− x < y\).