apply the distributive property and then simplify x-2/x squared -4 + 1/x+2 = 1/x-2

To solve the equation, let's apply the distributive property and then simplify step by step:

Given equation:

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

### Step 1: Factor the denominator \(x^2 - 4\)

Notice that \(x^2 - 4\) is a difference of squares, which can be factored as:

\[
x^2 - 4 = (x - 2)(x + 2)
\]

So, the equation becomes:

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

### Step 2: Simplify the first term

In the first term \(\frac{x-2}{(x-2)(x+2)}\), the \(x-2\) cancels out:

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

Thus, the equation simplifies to:

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

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

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

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

This simplifies to:

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

### Step 4: Cross-multiply to solve for \(x\)

Cross-multiplying gives:

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

### Step 5: Distribute and solve

Distribute the terms on both sides:

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

Subtract \(x\) from both sides:

\[
x - 4 = 2
\]

Add 4 to both sides:

\[
x = 6
\]

### Final Answer:

The solution is \(x = 6\).

Would you like to discuss the steps further or have any questions?

Here are some related questions:
1. How can you apply the distributive property in more complex equations?
2. What are some other examples of the difference of squares?
3. How does cross-multiplication work with different types of fractions?
4. What are the steps to factor quadratic expressions in general?
5. How do we handle potential restrictions on the variable \(x\) (like division by zero)?

**Tip:** Always check for extraneous solutions, especially when dealing with rational expressions, by substituting the solution back into the original equation.

Learn how to simplify the rational expression (x-2)/(x^2-4) + 1/(x+2) = 1/(x-2) step by step. This problem involves applying the distributive property and simplifying rational expressions, suitable for high school students in Grades 9-12.

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