The problem asks us to express the given rational expression, $\frac{4x-9}{(x-2)(x-3)}$, as a sum of partial fractions. We assume the form $\frac{A}{x-2} + \frac{B}{x-3}$, where $A$ and $B$ are constants we need to find.

1. Problem Description

The problem asks us to express the given rational expression, $\frac{4x-9}{(x-2)(x-3)}$, as a sum of partial fractions. We assume the form $\frac{A}{x-2} + \frac{B}{x-3}$, where $A$ and $B$ are constants we need to find.

2. Solution Steps

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

First, we write

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

Then, we multiply both sides by $(x-2)(x-3)$ to get

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

Now, we solve for $A$ and $B$.

Let $x = 2$. Then $4(2) - 9 = A(2-3) + B(2-2)$, which gives $8 - 9 = -A$, so $-1 = -A$, and $A = 1$.

Let $x = 3$. Then $4(3) - 9 = A(3-3) + B(3-2)$, which gives $12 - 9 = B$, so $3 = B$, and $B = 3$.

Therefore,

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

3. Final Answer

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