The problem is to decompose the rational function $\frac{4x-9}{(x-2)(x-3)}$ into partial fractions.

1. Problem Description

The problem is to decompose the rational function $\frac{4x-9}{(x-2)(x-3)}$ into partial fractions.

2. Solution Steps

We want to decompose $\frac{4x-9}{(x-2)(x-3)}$ into the form $\frac{A}{x-2} + \frac{B}{x-3}$.

So, we have

$\frac{4x-9}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$.

Multiplying both sides by $(x-2)(x-3)$, we get

$4x-9 = A(x-3) + B(x-2)$.

We can solve for $A$ and $B$ by substituting suitable values of $x$.

Let $x=2$. Then

$4(2) - 9 = A(2-3) + B(2-2)$

$8 - 9 = A(-1) + 0$

$-1 = -A$

$A = 1$.

Let $x=3$. Then

$4(3) - 9 = A(3-3) + B(3-2)$

$12 - 9 = 0 + B(1)$

$3 = B$

$B = 3$.

Thus, we have

$\frac{4x-9}{(x-2)(x-3)} = \frac{1}{x-2} + \frac{3}{x-3}$.

3. Final Answer

$\frac{1}{x-2} + \frac{3}{x-3}$