Find the largest integral $x$ which satisfies the following inequality:
$\frac{x + 4}{x^{2} - 9} - \frac{2}{x + 3} < \frac{4 x}{3 x - x^{2}}$

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Solution

$\frac{x + 4}{x^{2} - 9} - \frac{2}{x + 3} < \frac{4 x}{3 x - x^{2}}$
$\frac{x + 4}{(x - 3) (x + 3)} - \frac{2}{(x + 3)} + \frac{4 x}{x (x - 3)} < 0$
$\frac{x^{2} + 4 x - 2 x^{2} + 6 x + 4 x^{2} + 12 x}{x (x - 3) (x + 3)} < 0$
$\frac{3 x^{2} + 20 x}{x (x - 3) (x + 3)} < 0$
$\frac{x (3 x + 20)}{x (x - 3) (x + 3)} < 0$
$x \neq 0$
$\frac{3 x + 20}{(x - 3) (x + 3)} < 0$
$x \in (- \infty, - \frac{20}{3}) \cup (- 3, 3) - {0}$
The largest integral of x that satisfies the above inequality is $2$ .

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