2/x²-1 - 1/x-1 = 1/2 It's a rational equations 

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.

Learn how to solve the rational equation 2/x² - 1 - 1/(x - 1) = 1/2 with a detailed step-by-step solution. This problem covers key algebraic concepts, including rational equations, factoring, and cross-multiplication. Suitable for students in Grades 10-12.

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