check for extraneous solutions: $\quad \frac{x}{x-2}-\frac{1}{x-4}=-\frac{4}{x^{2}-6 x+8} \quad(7$ pts) \[ \frac{1}{x-4}+\frac{4}{x^{2}-6 x+8} \]

Step 1 ：Find the common denominator of the fractions on the left side of the equation: \((x-2)(x-4)\)

Step 2 ：Rewrite the fractions with the common denominator: \(\frac{x(x-4)}{(x-2)(x-4)} - \frac{(x-2)}{(x-2)(x-4)} + \frac{4}{(x-2)(x-4)} = 0\)

Step 3 ：Simplify the equation: \(\frac{x^2-4x-x^2+2x+4}{(x-2)(x-4)} = 0\)

Step 4 ：Solve for x: \(x-2 = 0\) which gives \(x = 3\)

Step 5 ：Check for extraneous solutions: None

Step 6 ：Final Answer: \(\boxed{3}\)