Let's solve the equation step by step. The given equation is:

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

### Step 1: Identify a common denominator

The left-hand side of the equation has two fractions, \(\frac{1}{x - 6}\) and \(\frac{1}{x - 2}\). Their common denominator would be \((x - 6)(x - 2)\).

We can rewrite the left-hand side as:
\[
\frac{(x - 2) + (x - 6)}{(x - 6)(x - 2)} = \frac{x - 2 + x - 6}{(x - 6)(x - 2)} = \frac{2x - 8}{(x - 6)(x - 2)}
\]

### Step 2: Simplify the right-hand side

The right-hand side is \(\frac{4}{x^2 - 8x + 12}\). Let's factor the quadratic expression \(x^2 - 8x + 12\):
\[
x^2 - 8x + 12 = (x - 6)(x - 2)
\]
So the right-hand side becomes:
\[
\frac{4}{(x - 6)(x - 2)}
\]

### Step 3: Set up the equation

Now that both sides have the same denominator, we can set the numerators equal to each other:
\[
2x - 8 = 4
\]

### Step 4: Solve for \(x\)

Solve the equation \(2x - 8 = 4\):
\[
2x = 4 + 8 = 12
\]
\[
x = \frac{12}{2} = 6
\]

### Step 5: Check for extraneous solutions

Since \(x = 6\) makes the denominators in the original equation undefined (because \(x - 6 = 0\)), it is an extraneous solution. Therefore, **there is no solution** to the equation.

Let me know if you'd like any further details or explanations.

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Here are 5 related questions you might explore:

1. What are extraneous solutions, and why do they arise?
2. How can you solve rational equations with multiple fractions?
3. What are the steps to factor quadratic expressions like \(x^2 - 8x + 12\)?
4. How do common denominators help in solving fractional equations?
5. What is the significance of checking for undefined values in rational equations?

**Tip:** Always check for values that make any denominator zero when solving equations with fractions!

Explore how to solve the rational equation (1/x-6)+(1/x-2)==4/x^2-8x+12 using common denominators and factoring quadratic expressions. This step-by-step solution includes key concepts like solving for extraneous solutions, making it ideal for high school students in Grades 10-12.

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