#### Answer

$=\frac{c^2-14c+4}{(3c-1)(c-2)}$

#### Work Step by Step

Removing factors both in the numerator and the denominator:
$\frac{c}{3c-1}-\frac{4}{c-2}$
$=\frac{c(c-2)}{(3c-1)(c-2)}-\frac{4(3c-1)}{(c-2)(3c-1)}$
$=\frac{c^2-2c}{(3c-1)(c-2)}-\frac{12c-4}{(3c-1)(c-2)}$
$=\frac{c^2-14c+4}{(3c-1)(c-2)}$