To solve the rational equation:

$\frac{3}{x} + \frac{2}{x - 4} = \frac{x - 2}{x - 4}$

Step 1: Identify the restrictions

Before solving, identify the restrictions by finding the values of $x$ that would make the denominators zero.

• For $x$, the denominator in $\frac{3}{x}$, we must have $x \neq 0$.
• For $x - 4$, the denominator in $\frac{2}{x - 4}$ and $\frac{x - 2}{x - 4}$, we must have $x \neq 4$.

So, $x \neq 0$ and $x \neq 4$ are the restrictions.

Step 2: Multiply through by the common denominator

The common denominator between the terms is $x(x - 4)$. Multiply both sides of the equation by this to eliminate the denominators:

$x(x - 4) \left( \frac{3}{x} + \frac{2}{x - 4} \right) = x(x - 4) \left( \frac{x - 2}{x - 4} \right)$

This simplifies as follows:

$(x - 4) \cdot 3 + x \cdot 2 = x(x - 2)$

$3x - 12 + 2x = x^2 - 2x$

Step 3: Simplify the equation

Now, combine like terms:

$5x - 12 = x^2 - 2x$

Move all terms to one side of the equation:

$0 = x^2 - 2x - 5x + 12$

$x^2 - 7x + 12 = 0$

Step 4: Solve the quadratic equation

Now, solve the quadratic equation $x^2 - 7x + 12 = 0$. We can factor this quadratic:

$(x - 3)(x - 4) = 0$

Thus, the solutions are:

$x = 3 \quad \text{or} \quad x = 4$

Step 5: Check for extraneous solutions

Recall that $x \neq 4$ is a restriction. Therefore, $x = 4$ is an extraneous solution and must be discarded.

The only valid solution is $x = 3$.

Final Answer:

The solution set is:

$\{ 3 \}$

Would you like further details or have any questions?

Here are 5 related questions to explore:

1. How do you factor quadratic equations with different methods?
2. What happens when rational equations have no solutions?
3. How do restrictions affect the solution of rational equations?
4. Can you explain how to clear denominators in rational equations?
5. How do you handle extraneous solutions in general?

Tip: Always check for extraneous solutions by considering the restrictions from the original denominators!