If the replacement set is the set of real numbers, solve : $\frac{x + 3}{8} < \frac{x - 3}{5}$

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Solution

$\frac{x + 3}{8} < \frac{x - 3}{5}$
$5 x + 15 < 8 x - 24$
$39 < 3 x$
$x > 13 \Rightarrow x ε (13, \infty)$

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