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

Answer

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Answer

Final Answer: $\boxed{3}$

Steps

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: $\boxed{3}$