Originally Posted by

**MarkFL2** My way is similar to that of **earboth**:

$\displaystyle \frac{x}{x-2}>5$

$\displaystyle \frac{x}{x-2}-5>0$

$\displaystyle \frac{x}{x-2}-\frac{5(x-2)}{x-2}>0$

$\displaystyle \frac{x-5x+10}{x-2}>0$

$\displaystyle \frac{10-4x}{x-2}>0$

$\displaystyle \frac{2(5-2x)}{x-2}>0$

Equating the numerator and denominator to zero, we find two critical numbers:

$\displaystyle x=2,\,\frac{5}{2}$

Dividing the real number line at these critical value gives us 3 intervals. Since neither of the critical numbers is a root of even multiplcity, we know the sign of the expression will alternate across the intervals. We pick a test number from the first (leftmost) interval:

$\displaystyle (-\infty,2)$ test value $\displaystyle x=0$ expression is negative, so our solution will be:

$\displaystyle \left(2,\frac{5}{2} \right)$

We use an open interval since the inequality is strict. If it were weak, we would still have to exclude 2 to avoid division by zero.