1/(x - 6) + x/(x - 2) = 4/(x ^ 2 - 8x + 12) rational equation

To solve the rational equation:

\[
\frac{1}{x - 6} + \frac{x}{x - 2} = \frac{4}{x^2 - 8x + 12}
\]

we first notice that the expression on the right side has a quadratic denominator. We can factor the quadratic expression:

\[
x^2 - 8x + 12 = (x - 6)(x - 2).
\]

Thus, the equation becomes:

\[
\frac{1}{x - 6} + \frac{x}{x - 2} = \frac{4}{(x - 6)(x - 2)}.
\]

To solve this, we find a common denominator for all terms, which is \((x - 6)(x - 2)\). Rewriting each fraction with this common denominator gives:

\[
\frac{(x - 2) + x(x - 6)}{(x - 6)(x - 2)} = \frac{4}{(x - 6)(x - 2)}.
\]

Simplify the numerator:

\[
(x - 2) + x(x - 6) = x - 2 + x^2 - 6x = x^2 - 5x - 2.
\]

So, the equation becomes:

\[
\frac{x^2 - 5x - 2}{(x - 6)(x - 2)} = \frac{4}{(x - 6)(x - 2)}.
\]

Since the denominators are the same, we equate the numerators:

\[
x^2 - 5x - 2 = 4.
\]

Solve for \(x\) by rearranging:

\[
x^2 - 5x - 6 = 0.
\]

Factor the quadratic equation:

\[
(x - 6)(x + 1) = 0.
\]

Thus, the solutions are:

\[
x = 6 \quad \text{or} \quad x = -1.
\]

### Check for Extraneous Solutions

1. For \(x = 6\):

   Substituting \(x = 6\) into the original denominators \((x - 6)\) and \((x - 2)\):

   \[
   x - 6 = 6 - 6 = 0.
   \]

   This makes the denominator zero, which is undefined. Thus, \(x = 6\) is an extraneous solution.

2. For \(x = -1\):

   Substituting \(x = -1\):

   \[
   x - 6 = -1 - 6 = -7 \neq 0, \quad x - 2 = -1 - 2 = -3 \neq 0.
   \]

   The denominators are non-zero, so \(x = -1\) is a valid solution.

### Final Answer

The solution to the equation is:

\[
\boxed{x = -1}.
\]

Would you like further details or have any questions?

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Here are 5 related questions for further exploration:

1. How do you solve other rational equations with more complex denominators?
2. What are extraneous solutions, and why do they occur in rational equations?
3. How can you graphically represent the solution to this rational equation?
4. How do you find the least common denominator for three or more rational expressions?
5. What are some real-world applications of solving rational equations?

**Tip:** Always check for extraneous solutions when solving rational equations, as they can result from restrictions in the domain.

Learn how to solve the rational equation 1/(x - 6) + x/(x - 2) = 4/(x^2 - 8x + 12) with detailed step-by-step solutions. Understand the concept of rational equations, factoring quadratic denominators, and identifying extraneous solutions. Suitable for Grades 10-12.

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