Solve the rational equation. -4x/x-1 + 4/x+1 = -8/x²-1

We are given the rational equation:

\[
\frac{-4x}{x-1} + \frac{4}{x+1} = \frac{-8}{x^2 - 1}
\]

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

Notice that \(x^2 - 1\) is a difference of squares:

\[
x^2 - 1 = (x - 1)(x + 1)
\]

So the equation becomes:

\[
\frac{-4x}{x-1} + \frac{4}{x+1} = \frac{-8}{(x-1)(x+1)}
\]

### Step 2: Find the least common denominator (LCD).

The least common denominator (LCD) of the fractions is \((x-1)(x+1)\). We will rewrite all terms with this common denominator:

- The first term: \(\frac{-4x}{x-1}\) can be rewritten as:

\[
\frac{-4x(x+1)}{(x-1)(x+1)} = \frac{-4x^2 - 4x}{(x-1)(x+1)}
\]

- The second term: \(\frac{4}{x+1}\) can be rewritten as:

\[
\frac{4(x-1)}{(x-1)(x+1)} = \frac{4x - 4}{(x-1)(x+1)}
\]

Now the equation becomes:

\[
\frac{-4x^2 - 4x}{(x-1)(x+1)} + \frac{4x - 4}{(x-1)(x+1)} = \frac{-8}{(x-1)(x+1)}
\]

### Step 3: Combine the fractions on the left-hand side.

Since the denominators are the same, we can combine the numerators:

\[
\frac{-4x^2 - 4x + 4x - 4}{(x-1)(x+1)} = \frac{-8}{(x-1)(x+1)}
\]

Simplifying the numerator:

\[
\frac{-4x^2 - 4}{(x-1)(x+1)} = \frac{-8}{(x-1)(x+1)}
\]

### Step 4: Solve the equation.

Since the denominators are the same, we can equate the numerators:

\[
-4x^2 - 4 = -8
\]

Simplify this equation:

\[
-4x^2 = -8 + 4
\]

\[
-4x^2 = -4
\]

Divide both sides by \(-4\):

\[
x^2 = 1
\]

### Step 5: Solve for \(x\).

Taking the square root of both sides:

\[
x = \pm 1
\]

### Step 6: Check for extraneous solutions.

We need to check if either solution makes the denominator zero. If \(x = 1\), then \(x - 1 = 0\), which makes the denominator zero in the original equation, so \(x = 1\) is an extraneous solution.

If \(x = -1\), then \(x + 1 = 0\), which also makes the denominator zero, so \(x = -1\) is another extraneous solution.

### Conclusion:

There are **no solutions** to the given equation because both \(x = 1\) and \(x = -1\) result in undefined expressions.

---

Would you like more details, or do you have any questions?

Here are five related questions to expand on this topic:
1. How do you find the least common denominator for rational equations?
2. What are extraneous solutions, and how do you identify them?
3. How do you simplify expressions with differences of squares?
4. Can rational equations have no solutions? Under what conditions?
5. What are the key steps to solving rational equations efficiently?

**Tip**: Always check for extraneous solutions when solving rational equations to avoid incorrect answers.

Learn how to solve a complex rational equation involving the difference of squares. This step-by-step solution includes factoring, finding the least common denominator, and identifying extraneous solutions. Ideal for students in Grades 9-11.

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