## Algebra 1: Common Core (15th Edition)

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