$\displaystyle

\frac{8x+9}{x+3} > \frac{x+3}{x-1}

$

I forgot how to solve these, don't you multiply each side by each's side denominator?

$\displaystyle

\frac{(8x+9)(x-1)}{x+3} - \frac{(x+3)(x+3)}{x-1} > 0

$

$\displaystyle

\frac{(8x^2+x-9)-(x^2+6x+9)}{(x+3)(x-1)}

$

$\displaystyle

\frac{7x^2-5x-18}{(x+3)(x-1)}

$

Thanks!