The problem requires us to perform a partial fraction decomposition on the rational expression $\frac{4x-9}{(x-2)(x-3)}$. This means we need to find constants $A$ and $B$ such that $\frac{4x-9}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$.

1. Problem Description

The problem requires us to perform a partial fraction decomposition on the rational expression $\frac{4x-9}{(x-2)(x-3)}$. This means we need to find constants $A$ and $B$ such that $\frac{4x-9}{(x-2)(x-3)} = \frac{A}{x-2} + \frac{B}{x-3}$.

2. Solution Steps

First, we write the given rational expression as a sum of two fractions with unknown constants $A$ and $B$:

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

Next, we multiply both sides of the equation by $(x-2)(x-3)$ to eliminate the denominators:

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

Now we can solve for $A$ and $B$ by choosing convenient values for $x$.

Let $x=2$. Then we have:

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

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

$$-1 = -A$$

$$A = 1$$

Let $x=3$. Then we have:

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

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

$$3 = B$$

$$B = 3$$

Therefore, we have $A=1$ and $B=3$. We can now write the partial fraction decomposition:

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

3. Final Answer

The partial fraction decomposition of the given expression is:

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