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

Answer: (D)

If $y < 0$ , then $− x < y$

Answer 2

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 $− x < y$.