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Question:

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

Explanation:

Given, $\frac{\mathrm{x}}{\mathrm{y}}$ < $\frac{\mathrm{x}+3}{\mathrm{y}-3}$ 

⇒ $\frac{\mathrm{x}}{\mathrm{y}}$ - $\frac{\mathrm{x}+3}{\mathrm{y}-3}$ < 0

⇒ $\frac{\mathrm{x}(\mathrm{y}-3)-\mathrm{y}(\mathrm{x}+3)}{\mathrm{y}(\mathrm{y}-3)}$ < 0

⇒ $\frac{\mathrm{xy}-3x-\mathrm{yx}-3y}{\mathrm{y}(\mathrm{y}-3)}$ < 0

⇒ $\frac{-3(x+y)}{\mathrm{y}(\mathrm{y}-3)}$ < 0

⇒ $\frac{(x+y)}{\mathrm{y}(\mathrm{y}-3)}$ > 0   ...(1)

Now, checking options.

Option (a): If y > 10, y(y - 3) > 0
From (1): ⇒ (x + y) > 0
⇒ -x < y
Hence, option (b) is incorrect.

Options (b) and (c) are slightly difficult to check. We can come back to them if required.

Option (d): If y < 0, y(y - 3) > 0
From (1): ⇒ (x + y) > 0
⇒ -x < y
Hence, option (d) is correct.

Hence, option (d).