We are given the equation $\frac{3x+4}{x^2-3x+2} = \frac{A}{x-1} + \frac{B}{x-2}$ and we are asked to find the value of $A+B$.

1. Problem Description

We are given the equation $\frac{3x+4}{x^2-3x+2} = \frac{A}{x-1} + \frac{B}{x-2}$ and we are asked to find the value of $A+B$.

2. Solution Steps

First, factor the denominator of the left-hand side:

$x^2-3x+2 = (x-1)(x-2)$

So the equation becomes:

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

Multiply both sides by $(x-1)(x-2)$ to clear the denominators:

$3x+4 = A(x-2) + B(x-1)$

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

Let $x=1$:

$3(1)+4 = A(1-2) + B(1-1)$

$7 = A(-1) + B(0)$

$7 = -A$

$A = -7$

Let $x=2$:

$3(2)+4 = A(2-2) + B(2-1)$

$6+4 = A(0) + B(1)$

$10 = B$

$B = 10$

Therefore, $A+B = -7+10 = 3$

3. Final Answer

$A+B=3$