# Rational Inequality Problem

• May 9th 2010, 10:28 PM
desiderius1
Rational Inequality Problem
$
\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?

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

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

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

Thanks!
• May 9th 2010, 11:46 PM
sa-ri-ga-ma
$7x^2 - 5x - 18 >0$

Factorize this, you will get,
(7x+9)(x-2)>0
Now find the range of the x values which satisfies the above condition.