Answer
$\frac{-3}{x+2}+\frac{1}{x+3}+\frac{1}{2x-3}$
Work Step by Step
Step 1. Factor the denominator as $(x+2)(x+3)(2x-3)$
Step 2. Assume the end results as $\frac{A}{x+2}+\frac{B}{x+3}+\frac{C}{2x-3}$
Step 3. Combine the above functions as:
$\frac{A(x+3)(2x-3)+B(x+2)(2x-3)+C(x+2)(x+3)}{(x+2)(x+3)(2x-3)}=
=\frac{(2A+2B+C)x^2+(3A+B+5C)x+(-9A-6B+6C)}{(x+2)(x+3)(2x-3)}$
Step 4. Compare the above result with the original to set up the following system of equations:
\begin{cases} 2A+2B+C=-3 \\ 3A+B+5C=-3 \\ -9A-6B+6C=27 \end{cases}
Step 5. Multiple 3 to the first equation and add it to the last equation, also multiply 6 to the second equation and add it to the last equation, we get:
\begin{cases} -3A+9C=18 \\ 9A+36C=9 \end{cases} or \begin{cases} -A+3C=6 \\ A+4C=1 \end{cases}
Step 6. Use substitution to solve the above equations and get $A=-3, B=1, C=1$
Step 7. Conclusion: the end result is:
$\frac{-3}{x+2}+\frac{1}{x+3}+\frac{1}{2x-3}$
