Question

Solve the following inequality:
$\frac{7x-5}{8x+3}>4$

Answer 1

$\frac{7x-5}{8x+3}>4$

$\frac{7x-5}{8x+3}-4>0$

$\frac{7x-5-32x-12}{8x+3}>0$

$\frac{25x+17}{8x+3}<0$

$xϵ(\frac{-17}{25},\frac{-3}{8})$