Question:
$\frac{7 x-5}{8 x+3}>4$
Solution:
We have, $\frac{7 x-5}{8 x+3}>4$
$\Rightarrow \frac{7 x-5}{8 x+3}-4>0$
$\Rightarrow \frac{7 x-5-4(8 x+3)}{8 x+3}>0$
$\Rightarrow \frac{7 x-5-32 x-12}{8 x+3}>0$
$\Rightarrow \frac{-25 x-17}{8 x+3}>0$
$\Rightarrow \frac{25 x+17}{8 x+3}<0 \quad$ (Multiplying by $-1$ to make the coefficient of $x$ in the LHS positive)
$\mathrm{x} \in\left(\frac{-17}{25}, \frac{-3}{8}\right)$
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