Solve the equation 2/(x-2) = x/(x-2) - x/3 for x.

1. Identify the equation: The given equation is a rational equation: $$\frac{2}{x-2} = \frac{x}{x-2} - \frac{x}{3}$$
2. Determine restrictions: The denominators cannot be zero. Thus, $x-2 \neq 0$, which means $x \neq 2$. Also, $3 \neq 0$, which is always true. So, the only restriction is $x \neq 2$.
3. Find a common denominator: The denominators are $(x-2)$, $(x-2)$, and $3$. The least common denominator (LCD) is $3(x-2)$.
4. Multiply each term by the LCD: Multiply every term in the equation by $3(x-2)$ to eliminate the denominators:
$$3(x-2) \left( \frac{2}{x-2} \right) = 3(x-2) \left( \frac{x}{x-2} \right) - 3(x-2) \left( \frac{x}{3} \right)$$
5. Simplify the equation: Cancel out common factors in each term:
$$3 \cdot 2 = 3 \cdot x - (x-2) \cdot x$$
$$6 = 3x - (x^2 - 2x)$$
6. Distribute and combine like terms:
$$6 = 3x - x^2 + 2x$$
$$6 = -x^2 + 5x$$
7. Rearrange into a quadratic equation: Move all terms to one side to set the equation to zero:
$$x^2 - 5x + 6 = 0$$
8. Factor the quadratic equation: Find two numbers that multiply to $6$ and add to $-5$. These numbers are $-2$ and $-3$.
$$(x-2)(x-3) = 0$$
9. Solve for x: Set each factor equal to zero and solve:
$$x-2 = 0 \quad \text{or} \quad x-3 = 0$$
$$x = 2 \quad \text{or} \quad x = 3$$
10. Check for extraneous solutions: Recall the restriction from step 2: $x \neq 2$. Therefore, $x=2$ is an extraneous solution. The only valid solution is $x=3$.
11. Final Answer: The only valid solution is $x=3$.