Chapter 5, Systems of Equations and Inequalities - Section 5.3 - Partial Fractions - 5.3 Exercises - Page 462: 16

$\frac{x+6}{(x)(x+3)}=\frac{2}{x}+\frac{-1}{x+3}$

$\frac{x+6}{(x)(x+3)}=\frac{A}{x}+\frac{B}{x+3}$, $x+6=A(x+3)+Bx$, $x+6=Ax+3A+Bx$, $x+6=(A+B)x+3A$, $(A+B)x=x$, $A+B=1$ and $3A=6, A=2$ thus, substituting back into the equation, $2+B=1, B=-1$ $\frac{x+6}{(x)(x+3)}=\frac{2}{x}+\frac{-1}{x+3}$