To solve the rational equation:

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

Let's proceed step by step:

Step 1: Factor the denominators

We know that $x^2 - 1$ is a difference of squares, so we can factor it: $x^2 - 1 = (x - 1)(x + 1)$ This allows us to rewrite the equation as: $\frac{2}{(x - 1)(x + 1)} - \frac{1}{x - 1} = \frac{1}{2}$

Step 2: Find a common denominator

The common denominator between the two fractions on the left-hand side is $(x - 1)(x + 1)$. We'll rewrite the second fraction with this denominator: $\frac{2}{(x - 1)(x + 1)} - \frac{1(x + 1)}{(x - 1)(x + 1)} = \frac{1}{2}$ This simplifies to: $\frac{2 - (x + 1)}{(x - 1)(x + 1)} = \frac{1}{2}$

Step 3: Simplify the numerator

Simplify the numerator on the left-hand side: $2 - (x + 1) = 2 - x - 1 = 1 - x$ Now the equation becomes: $\frac{1 - x}{(x - 1)(x + 1)} = \frac{1}{2}$

Step 4: Cross-multiply to eliminate the fractions

Cross-multiply to solve for $x$: $2(1 - x) = (x - 1)(x + 1)$ Expand both sides: $2 - 2x = x^2 - 1$

Step 5: Rearrange the equation into standard quadratic form

Move all terms to one side of the equation: $x^2 + 2x - 3 = 0$

Step 6: Factor the quadratic equation

Factor the quadratic equation: $(x + 3)(x - 1) = 0$

Step 7: Solve for $x$

Set each factor equal to zero: $x + 3 = 0 \quad \text{or} \quad x - 1 = 0$ $x = -3 \quad \text{or} \quad x = 1$

Step 8: Check for extraneous solutions

We must exclude any solutions that make the original denominators zero. In the original equation, $x = 1$ makes the denominators zero, so $x = 1$ is an extraneous solution.

Thus, the only valid solution is: $x = -3$

Final Answer

The solution to the equation is: $x = -3$

Would you like more details or have any questions?

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Here are 5 related questions you could explore:

1. How do you identify extraneous solutions in rational equations?
2. How would the process change if the equation had higher powers of $x$?
3. Can rational equations have no solutions? Why?
4. What are the consequences of cross-multiplying in equations with variables in the denominator?
5. How does factoring play a role in solving rational equations?

Tip: Always check for extraneous solutions when solving rational equations, as these can occur if the variable makes any denominator zero.