Let's solve the given rational equation step by step.

Problem:

$\frac{x}{x - 1} + \frac{1}{x - 2} = \frac{1}{x^2 - 3x + 2}$

Step 1: Factor the denominator on the right-hand side

The denominator on the right-hand side is $x^2 - 3x + 2$. We can factor it as:

$x^2 - 3x + 2 = (x - 1)(x - 2)$

So, the equation becomes:

$\frac{x}{x - 1} + \frac{1}{x - 2} = \frac{1}{(x - 1)(x - 2)}$

Step 2: Identify the Least Common Denominator (LCD)

The LCD of the denominators $(x - 1), (x - 2),$ and $(x - 1)(x - 2)$ is $(x - 1)(x - 2)$.

Step 3: Rewrite each fraction with the LCD

We rewrite all terms so that each has the same denominator $(x - 1)(x - 2)$:

• The first term becomes: $\frac{x}{x - 1} = \frac{x(x - 2)}{(x - 1)(x - 2)} = \frac{x^2 - 2x}{(x - 1)(x - 2)}$
• The second term becomes: $\frac{1}{x - 2} = \frac{1(x - 1)}{(x - 1)(x - 2)} = \frac{x - 1}{(x - 1)(x - 2)}$
• The third term is already: $\frac{1}{(x - 1)(x - 2)}$

So the equation now becomes:

$\frac{x^2 - 2x}{(x - 1)(x - 2)} + \frac{x - 1}{(x - 1)(x - 2)} = \frac{1}{(x - 1)(x - 2)}$

Step 4: Combine the terms on the left-hand side

Now, combine the two fractions on the left-hand side since they have the same denominator:

$\frac{(x^2 - 2x) + (x - 1)}{(x - 1)(x - 2)} = \frac{x^2 - x - 1}{(x - 1)(x - 2)}$

So the equation becomes:

$\frac{x^2 - x - 1}{(x - 1)(x - 2)} = \frac{1}{(x - 1)(x - 2)}$

Step 5: Cancel the denominators

Since the denominators are the same, we can cancel them out:

$x^2 - x - 1 = 1$

Step 6: Solve the quadratic equation

Now, solve the quadratic equation:

$x^2 - x - 1 = 1$

Subtract 1 from both sides:

$x^2 - x - 2 = 0$

Now, factor the quadratic equation:

$x^2 - x - 2 = (x - 2)(x + 1) = 0$

Set each factor equal to zero:

$x - 2 = 0 \quad \text{or} \quad x + 1 = 0$

So, $x = 2$ or $x = -1$.

Step 7: Check for extraneous solutions

We must check if any solutions make the original denominators zero, because division by zero is undefined.

• For $x = 2$, the denominator $x - 2$ becomes zero. Therefore, $x = 2$ is not a valid solution.
• For $x = -1$, none of the denominators $x - 1$ or $x - 2$ becomes zero. Therefore, $x = -1$ is a valid solution.

Final Answer:

The only solution is $\boxed{x = -1}$.

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Would you like to see details or ask any questions?

Here are 5 relative questions for further understanding:

1. How do we factor quadratic equations like $x^2 - 3x + 2$?
2. Why is the LCD important in solving rational equations?
3. How can we identify extraneous solutions in rational equations?
4. What method can we use if the quadratic does not factor easily?
5. Can we solve this equation graphically to confirm the solution?

Tip: Always check your solutions by substituting them back into the original equation to ensure they don’t cause undefined expressions (like division by zero).