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The intervals...

Question

The interval(s) of $x$ which satisfies the inequality $\frac{1}{x + 2} < \frac{1}{3 x + 7}$ is/are

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Solution

$\frac{1}{x + 2} < \frac{1}{3 x + 7}$
$\Rightarrow \frac{1}{x + 2} - \frac{1}{3 x + 7} < 0$
$\Rightarrow \frac{3 x + 7 - x - 2}{(x + 2) (3 x + 7)} < 0$
$\Rightarrow \frac{2 x + 5}{(x + 2) (3 x + 7)} < 0$

Critical points are $- \frac{5}{2}, - \frac{7}{3}, - 2$

Therefore, $x \in (- \infty, - \frac{5}{2}) \cup (- \frac{7}{3}, - 2)$

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