We are asked to solve the equation $\frac{1}{x-3} + \frac{1}{x-4} = \frac{5}{x^2 - 7x + 12}$ for $x$.

1. Problem Description

We are asked to solve the equation $\frac{1}{x-3} + \frac{1}{x-4} = \frac{5}{x^2 - 7x + 12}$ for $x$.

2. Solution Steps

First, factor the denominator on the right side of the equation:

$x^2 - 7x + 12 = (x-3)(x-4)$

So the equation becomes:

$\frac{1}{x-3} + \frac{1}{x-4} = \frac{5}{(x-3)(x-4)}$

Next, find a common denominator for the left side of the equation:

$\frac{1(x-4)}{(x-3)(x-4)} + \frac{1(x-3)}{(x-4)(x-3)} = \frac{5}{(x-3)(x-4)}$

Combine the fractions on the left:

$\frac{x-4+x-3}{(x-3)(x-4)} = \frac{5}{(x-3)(x-4)}$

Simplify the numerator on the left:

$\frac{2x-7}{(x-3)(x-4)} = \frac{5}{(x-3)(x-4)}$

Since the denominators are equal, we can equate the numerators:

$2x - 7 = 5$

Add 7 to both sides of the equation:

$2x = 12$

Divide both sides by 2:

$x = 6$

We need to check if $x=6$ is a valid solution by ensuring it doesn't make any of the denominators zero.

$x-3 = 6-3 = 3 \neq 0$

$x-4 = 6-4 = 2 \neq 0$

$x^2 - 7x + 12 = (6)^2 - 7(6) + 12 = 36 - 42 + 12 = 6 \neq 0$

Since none of the denominators are zero when $x=6$, the solution is valid.

3. Final Answer

$x = 6$