Answer
$\dfrac{x}{(x-3)(x+2)}=\dfrac{3}{5(x-3)}+\dfrac{2}{5(x+2)}$
Work Step by Step
We are given the fraction:
$\dfrac{x}{(x-3)(x+2)}$
The denominator is factored, therefore we can write the partial fraction decomposition:
$\dfrac{x}{(x-3)(x+2)}=\dfrac{A}{x-3}+\dfrac{B}{x+2}$
Multiply each terms by the least common denominator $(x-3)(x+2)$:
$(x-3)(x+2)\cdot\dfrac{x}{(x-3)(x+2)}=(x-3)(x+2)\cdot\dfrac{A}{x-3}+(x-3)(x+2)\cdot\dfrac{B}{x+2}$
$x=A(x+2)+B(x-3)$
$x=Ax+2A+Bx-3B$
$x=(A+B)x+(2A-3B)$
Identify the coefficients:
$\begin{cases}
A+B=1\\
2A-3B=0
\end{cases}$
Solve the system:
$\begin{cases}
-2A-2B==-2\\
2A-3B=0
\end{cases}$
$-2A-2B+2A-3B=-2+0$
$-5B=-2$
$B=\dfrac{2}{5}$
$A+B=1$
$A+\dfrac{2}{5}=1$
$A=1-\dfrac{2}{5}$
$A=\dfrac{3}{5}$
The partial fraction decomposition is:
$\dfrac{x}{(x-3)(x+2)}=\dfrac{\dfrac{3}{5}}{x-3}+\dfrac{\dfrac{2}{5}}{x+2}$
$\dfrac{x}{(x-3)(x+2)}=\dfrac{3}{5(x-3)}+\dfrac{2}{5(x+2)}$
