Question

$\frac{x + 4}{x -3} < 2$ is satisfied when $x$ satisfies

• A. $\left(- \infty,\, 3\right) \cup \left(10,\, \infty\right)$
• B. $(3,\,10)$
• C. $\left(- \infty,\, 3\right) \cup [10,\, \infty)$
• D. None of these

Answer 1

Answer: (A)

$\left(- \infty,\, 3\right) \cup \left(10,\, \infty\right)$

Answer 2

We have, $\frac{x + 4}{x -3} < 2$ or $\frac{x + 4}{x -3} -2 < 0$ or $\quad\frac{-x + 10}{x -3} < 0$ or $\frac{x - 10}{x -3} > 0$ $\Rightarrow\quad$ \\{x - 10 > 0 and x-3>0\\} or \\{ x - 10<0 and x-3<0\\} \Rightarrow \quad \\{x > 10 and x > 3\\} or \\{x < 10 and x < 3\\} $\Rightarrow \quad x\in \left(-\infty,\,3\right) \cup \left(10,\, \infty\right)$