Answer
$\displaystyle \frac{x}{(x-3)(x-2)}=\frac{3}{x-3}-\frac{2}{x-2}$
Work Step by Step
Both factors of th$e$ denominator are distinct linear factors...
see p.541. Decomposition with Distinct Linear Factors In the Denominator:
Include one partial fraction with a $constant$ numerator for each distinct linear factor in the denominator.
Setup:
$\displaystyle \frac{x}{(x-3)(x-2)}=\frac{A}{x-3}+\frac{B}{x-2}\quad /\times(x-3)(x-2)\quad$
... multipliy with the LCD ...
$x=A(x-2)+B(x-3)$
$ x=Ax-2A+Bx-3B\qquad$ /group like terms...
$x+0=x(A+B)+(-2A-3B)$
Emphasize the 0 constant term on the LHS,
as we now equate the coefficients of like powers of x...
$\left\{\begin{array}{l}
1=A+B\\
0=-2A-3B
\end{array}\right.\qquad $solve the system
Multiply equation 1. with 3 and add the equations:
$3+0=3A-2A+3B-3B$
$3=A$
back substitute: $1=3+B$
$B=-2$
Rewrite the setup decomposition with $A=3, B=-2$
$\displaystyle \frac{x}{(x-3)(x-2)}=\frac{3}{x-3}-\frac{2}{x-2}$
