Solve for x :
`(x - 1)/(2x + 1) + (2x + 1)/(x -1) = 2` , where `x ne - (1)/(2) , 1`

Answer

To solve the equation

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

we will follow these steps:

### Step 1: Find a common denominator
The common denominator for the fractions is \((2x + 1)(x - 1)\).

### Step 2: Rewrite the equation
Multiply both sides of the equation by the common denominator:

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

### Step 3: Expand both sides
Expanding the left side:

\[ (x - 1)^2 + (2x + 1)^2 = (x^2 - 2x + 1) + (4x^2 + 4x + 1) = 5x^2 + 2x + 2. \]

Expanding the right side:

\[ 2(2x + 1)(x - 1) = 2(2x^2 - 2x + x - 1) = 2(2x^2 - x - 1) = 4x^2 - 2x - 2. \]

### Step 4: Set the equation to zero
Now, we set the equation to zero:

\[ 5x^2 + 2x + 2 - (4x^2 - 2x - 2) = 0. \]

This simplifies to:

\[ 5x^2 + 2x + 2 - 4x^2 + 2x + 2 = 0, \]
\[ x^2 + 4x + 4 = 0. \]

### Step 5: Factor the quadratic equation
The quadratic \(x^2 + 4x + 4\) can be factored as:

\[ (x + 2)(x + 2) = 0, \]
or simply:

\[ (x + 2)^2 = 0. \]

### Step 6: Solve for x
Setting the factor equal to zero gives:

\[ x + 2 = 0 \implies x = -2. \]

### Step 7: Check for extraneous solutions
We need to ensure that \(x = -2\) does not violate the conditions \(x \neq -\frac{1}{2}\) and \(x \neq 1\). Since \(-2\) does not equal either of these values, it is a valid solution.

### Final Answer
Thus, the solution to the equation is:

\[ \boxed{-2}. \]