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

Given equation:

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

Step 1: Find the Least Common Denominator (LCD)

The denominators in the equation are:

1. $x^2 - 1$
2. $x - 1$
3. 2

First, factor $x^2 - 1$: $x^2 - 1 = (x - 1)(x + 1)$ So, the LCD is the product of all distinct factors: $\text{LCD} = 2(x - 1)(x + 1)$

Step 2: Multiply both sides by the LCD

Multiply the entire equation by the LCD $2(x - 1)(x + 1)$: $2(x - 1)(x + 1) \left( \frac{2}{(x - 1)(x + 1)} - \frac{1}{x - 1} \right) = 2(x - 1)(x + 1) \left( \frac{1}{2} \right)$

Step 3: Apply the Distributive Property and simplify

Simplifying each term:

• First term: $2(x - 1)(x + 1) \cdot \frac{2}{(x - 1)(x + 1)} = 4$
• Second term: $2(x - 1)(x + 1) \cdot \frac{1}{x - 1} = 2(x + 1)$
• Right side: $2(x - 1)(x + 1) \cdot \frac{1}{2} = (x - 1)(x + 1) = x^2 - 1$

So, the equation becomes: $4 - 2(x + 1) = x^2 - 1$

Step 4: Find all possible values of $x$

Simplify the equation: $4 - 2x - 2 = x^2 - 1$ $2 - 2x = x^2 - 1$ Rearrange all terms to one side: $0 = x^2 + 2x - 3$ Solve the quadratic equation: $x^2 + 2x - 3 = 0$ Factor the quadratic: $(x + 3)(x - 1) = 0$ Thus, the possible solutions are: $x = -3 \quad \text{or} \quad x = 1$

Step 5: Check for extraneous roots

Substitute $x = -3$ into the original equation: $\frac{2}{(-3)^2 - 1} - \frac{1}{-3 - 1} = \frac{1}{2}$ $\frac{2}{9 - 1} - \frac{1}{-4} = \frac{1}{2}$ $\frac{2}{8} + \frac{1}{4} = \frac{1}{2}$ $\frac{1}{4} + \frac{1}{4} = \frac{1}{2}$ True. So, $x = -3$ is a valid solution.

Now, substitute $x = 1$: $\frac{2}{1^2 - 1} - \frac{1}{1 - 1} = \frac{1}{2}$ The second term involves division by zero, which is undefined. So, $x = 1$ is an extraneous solution and must be rejected.

Final Answer:

The only solution is: $x = -3$

Would you like to go over any details, or have further questions?

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Here are 5 related questions to extend your understanding:

1. How do we find the LCD of more complex rational expressions?
2. What are the steps to solving equations with multiple extraneous solutions?
3. How can we factor quadratics like $x^2 + 2x - 3$?
4. What role do restrictions on domain play in rational equations?
5. How do we handle equations where all potential solutions are extraneous?

Tip: Always check for extraneous solutions by substituting them back into the original equation, especially when there are restrictions like division by zero!