Chapter 6 - Section 6.2 - Adding and Subtracting Rational Expressions - Exercise Set - Page 427: 39

$\frac{13}{(y-3)(y-2)}$

$y^2-5y+6=y^2-2y-3y+6=y(y-2)-3(y-3)=(y-3)(y-2)$, so LCD$=(y-3)(y-2)$, thus $\frac{3y+7}{y^2-5y+6}-\frac{3}{y-3}=\frac{3y+7}{(y-3)(y-2)}-\frac{3(y-2)}{(y-3)(y-2)}=\frac{3y+7-3y+6}{(y-3)(y-2)}=\frac{13}{(y-3)(y-2)}$