To solve the equation:
[tex]\[
\frac{x}{x-2}+\frac{1}{x-6}=\frac{4}{x^2-8x+12}
\][/tex]
we follow these steps:
### Step 1: Simplify the Denominator
First, note that the denominator on the right-hand side can be factored:
[tex]\[
x^2 - 8x + 12 = (x - 2)(x - 6)
\][/tex]
Thus, the equation becomes:
[tex]\[
\frac{x}{x-2} + \frac{1}{x-6} = \frac{4}{(x-2)(x-6)}
\][/tex]
### Step 2: Combine the Left-hand Side
To combine the fractions on the left-hand side over a common denominator:
[tex]\[
\frac{x}{x-2} + \frac{1}{x-6}
\][/tex]
we use the common denominator [tex]\((x-2)(x-6)\)[/tex]:
[tex]\[
\frac{x(x-6) + 1(x-2)}{(x-2)(x-6)} = \frac{x^2 - 6x + x - 2}{(x-2)(x-6)} = \frac{x^2 - 5x - 2}{(x-2)(x-6)}
\][/tex]
### Step 3: Equate the Numerators
Since the denominators are now the same, we equate the numerators of both sides:
[tex]\[
x^2 - 5x - 2 = 4
\][/tex]
Simplifying this, we get:
[tex]\[
x^2 - 5x - 2 - 4 = 0
\][/tex]
[tex]\[
x^2 - 5x - 6 = 0
\][/tex]
### Step 4: Solve the Quadratic Equation
Factor or use the quadratic formula to solve:
[tex]\[
x^2 - 5x - 6 = 0 \implies (x - 6)(x + 1) = 0
\][/tex]
Thus, we find the solutions:
[tex]\[
x - 6 = 0 \quad \text{or} \quad x + 1 = 0
\][/tex]
[tex]\[
x = 6 \quad \text{or} \quad x = -1
\][/tex]
### Step 5: Check for Extraneous Solutions
We must check if these solutions are valid within the original equation, specifically ensuring they don't cause an undefined expression:
1. For [tex]\(x = 6\)[/tex]:
Substitute [tex]\(x = 6\)[/tex]:
[tex]\[
\frac{6}{6-2} + \frac{1}{6-6} \quad \text{is undefined due to the term} \quad \frac{1}{0}
\][/tex]
Thus [tex]\(x = 6\)[/tex] is not a valid solution.
2. For [tex]\(x = -1\)[/tex]:
Substitute [tex]\(x = -1\)[/tex]:
[tex]\[
\frac{-1}{-1-2} + \frac{1}{-1-6} = \frac{-1}{-3} + \frac{1}{-7} = \frac{1}{3} - \frac{1}{7} = \frac{7-3}{21} = \frac{4}{21}
\][/tex]
Check the right-hand side for [tex]\(x = -1\)[/tex]:
[tex]\[
\frac{4}{1 - 8 + 12} = \frac{4}{5}
\][/tex]
Since both sides equal [tex]\(\frac{4}{21}\)[/tex], [tex]\(x = -1\)[/tex] is a valid solution.
### Conclusion
Thus, the correct solution to the equation is:
[tex]\[
x = -1
\][/tex]
