Find the largest integral $x$ which satisfies the following inequality:
$\frac{4 x + 19}{x + 5} < \frac{4 x - 17}{x - 3}$

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Solution

$\frac{4 x + 19}{x + 5} < \frac{4 x - 17}{x - 3}$
$\frac{4 x^{2} + 19 x - 12 x - 57 - 4 x^{2} + 17 x - 20 x + 85}{(x + 5) (x - 3)} < 0$
$\frac{4 x + 28}{(x + 5) (x - 3)} < 0$
$\frac{x + 7}{(x - 5) (x - 3)} < 0$
$x \in (- \infty, - 7) \cup (- 5, 3)$
Largest integral value of $x$ is $2$

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