jmc

We factor the denominator in the left-hand side to get $$\frac{Bx-11}{(x-2)(x-5)}=\frac{A}{x-2}+\frac{3}{x-5}.$$ We then multiply both sides by $(x-2)(x-5)$, to get $$Bx-11=A(x-5)+3(x-2).$$ We can solve for $B$ substituting a suitable value of $x$. For example, setting $x=5$, the equation becomes $5B-11=9$, so $B=4$. Then $$4x-11=A(x-5)+3(x-2).$$ Setting $x=2$, this equation becomes $-3=-3A$, so $A=1$. Therefore, $A+B=1+4=\boxed{5}$.